H.1)
\(\mathbf {f}\in C^1(\Omega ;\mathbb {R}^N)\) ;
H.2)
 \(\mathbf {f}(-\mathbf {y},-\mathbf {u})=-\mathbf {f}(\mathbf {y},\mathbf {u}),\)  for every \((\mathbf {y},\,\mathbf {u})\in \Omega \times \mathbb {R}^N\) ;
H.3)
 there exists a function \(\Phi \in C^2(\Omega )\)  such that
\(\mathbf {f}=\nabla _{\mathbf {u}}\Phi ,\qquad \Phi (\mathbf {y},\mathbf {u})=\kappa \frac{|\mathbf {u}|^p}{|\mathbf {y}|^{N+\alpha p}}+\Psi (\mathbf {y},\mathbf {u}),\quad {\mbox{ for every }}\, (\mathbf {y},\,\mathbf {u})\in \Omega ,\) 
where \(\kappa ,\, p,\,\alpha \)  are constants such that
\(\kappa >0,\qquad 0<\alpha <1,\qquad p\ge 2,\) 
and
\(& \Psi (\mathbf {y},\mathbf {0})=0\le \Psi (\mathbf {y},\mathbf {u}),\\&|\nabla _\mathbf {u}\Psi (\mathbf {y},\mathbf {u})|, |D^2_{\mathbf {u}}\Psi (\mathbf {y},\mathbf {u})|\le g(\mathbf {y}),\quad {\mbox{ for every }}\,(\mathbf {y},\,\mathbf {u})\in \Omega ,\) 

for some nonnegative function \(g\in L^2_{\rm {loc}}(\mathbb {R}^N)\) .


We recall that in the peridynamics model the
\(\mathbb {R}^N\) -valued function \(\mathbf {u}\)  models the displacement vector field.
With this respect, assumption REF 
can be seen as a counterpart of Newton's Third Law of Motion (the Action-Reaction Law).
Also, assumption REF  states that the material is hyperelastic (the linear elastic case corresponding to \(p=2\) 
and \(\Psi =0\) ,
and the hyperelasticity taking into account
nonlinear elastic responses of the material).
In the present paper we focus on the simplest case planned by the previous theory, namely we will deal with \(p=2\) 
(quadratic elastic energy) and \(N=1\)  which represents a linear one-dimensional nonlocal mechanical model
(higher dimensional cases can be taken into account as well, but the analysis
is obviously more transparent when \(N=1\) ).
We intend to exploit all the relevant analytical aspects encoded in this problem and compare them with their physical counterparts.
In this perspective (REF ) reduces to the study of the following Cauchy problem
\({\left\lbrace \begin{array}{ll}\rho \,u_{tt}=-2\,\kappa \displaystyle \int _{-\delta }^{\delta }\frac{u(t,x)-u(t,x-y)}{|y|^{1+2\,\alpha }}dy=: K(u),&\quad t>0,\,x\in \mathbb {R},\\[10pt]u(0,x)=v_0(x),&\quad x\in \mathbb {R},\\[5pt]u_t(0,x)=v_1(x),&\quad x\in \mathbb {R}\,,\end{array}\right.}\) 
where \(\delta \) , \(\kappa \)  and \(\rho \)  are positive real constants and \(0<\alpha <1\) .
As customary, the integral in (REF ) is interpreted
in the “principal value” sense to “average out” the singularity,
namely
\(\begin{split}\displaystyle \int _{-\delta }^{\delta }\frac{u(t,x)-u(t,x-y)}{|y|^{1+2\,\alpha }}\,dy&:=\lim _{\epsilon \rightarrow 0^+}\displaystyle \int _{(-\delta ,\delta )\setminus (-\epsilon ,\epsilon )}\frac{u(t,x)-u(t,x-y)}{|y|^{1+2\,\alpha }}\,dy\\&=\frac{1}{2}\lim _{\epsilon \rightarrow 0^+}\displaystyle \int _{(-\delta ,\delta )\setminus (-\epsilon ,\epsilon )}\frac{2u(t,x)-u(t,x+y)-u(t,x-y)}{|y|^{1+2\,\alpha }}\,dy\\&=\frac{1}{2}\displaystyle \int _{-\delta }^{\delta }\frac{2u(t,x)-u(t,x+y)-u(t,x-y)}{|y|^{1+2\,\alpha }}\,dy.\end{split}\) 
The evolution problem in (REF )
is explicitly solvable, according to the following result:
Theorem 1.1 
LetIn this paper, for simplicity,
unless differently specified, we will take
the initial data \(v_0\)  and \(v_1\) 
in the
Schwartz Space of smooth and rapidly decreasing functions
(more general settings can be treated similarly
with technical modifications). \(v_0,\,v_1\in \mathcal {S}(\mathbb {R})\)  and \(0<\alpha <1\) .
Then problem (REF ) has the unique solution
\(u:\mathbb {R}_+\times \mathbb {R}\rightarrow \mathbb {R}\)  given by
\(u(t,x)=\int _{\mathbb {R}} e^{-i\xi x}\left[\widehat{v_0}(\xi )\cos \left(\omega (\xi )\,t\right)+\frac{\widehat{v_1}(\xi )}{\omega (\xi )}\sin \left(\omega (\xi )\,t\right)\right] d\xi \,,\) 
where \(\widehat{v_0}(\xi )\)  and \(\widehat{v_1}(\xi )\)  represent the Fourier transformWe use here the nonunitary convention that
\( \widehat{v}(\xi ):=\frac{1}{2\pi }\int _\mathbb {R}v(x)\,e^{ix\xi }\,dx.\) 
In this way, the inversion formula reads
\( v(x)=\int _\mathbb {R}\widehat{v}(\xi )\,e^{-i\xi x}\,d\xi .\) 

of \(v_0(x)\)  and \(v_1(x)\) ,
and \(\omega : \mathbb {R}\rightarrow \mathbb {R}^+\)  is the dispersion relation defined by
\(\omega (\xi )=\left(\frac{2\kappa }{\rho \,\delta ^{2\alpha }}\,\int _{-1}^{1}\frac{1-\cos (\xi \delta z)}{|z|^{1+2\alpha }}dz\right)^{1/2}\,.\) 
Additionally,
\(\omega ^2(\xi )\le \frac{C\kappa }{\rho \,\delta ^{2\alpha }}\,\left[\frac{|\xi |^{2}\delta ^{2}}{1-\alpha }\,\min \left\lbrace \frac{1}{|\xi |^{2-2\alpha }\delta ^{2-2\alpha }},1\right\rbrace +\frac{1}{\alpha }\chi _{\left( 0,1\right)}\left(\frac{2}{|\xi |\delta }\right) \,\left( \frac{|\xi |^{2\alpha }\delta ^{2\alpha }}{2^{2\alpha }}-1\right)\right],\) 
for some constant \(C>0\) 
(independent of all the parameters involved in
problem (REF )).

The existence and uniqueness of the solution of (REF ) follow from [1]}.
Thus, to check (REF ), up to a superposition,
we seek a solution of the form
\(u(t,x)=e^{\pm (i\omega t+i \xi x)}.\) 
After substituting this expression into (REF ), we obtain the equation
\(\omega ^2=&\frac{2\kappa }{\rho }\int _{-\delta }^{\delta }\frac{1-e^{\mp i\xi y}}{|y|^{1+2\alpha }}dy\nonumber \\=&\frac{2\kappa }{\rho }\int _{-\delta }^{\delta }\frac{1-\cos ( \xi y)}{|y|^{1+2\alpha }}dy\mp \frac{2i\kappa }{\rho }\int _{-\delta }^{\delta }\frac{\sin (\xi y)}{|y|^{1+2\alpha }}dy\nonumber \\=&\frac{2\kappa }{\rho }\int _{-\delta }^{\delta }\frac{1-\cos ( \xi y)}{|y|^{1+2\alpha }}dy=\frac{2\kappa }{\rho }\,\delta ^{-2\alpha }\int _{-1}^{1}\frac{1-\cos (\xi \delta z)}{|z|^{1+2\alpha }}dz\,,\) 

which leads to the dispersion relation (REF ).
Introducing the functions
\(\alpha (\xi ):=\frac{1}{2}\left(\widehat{v_0}(\xi )+i\frac{\widehat{v_1}(\xi )}{\omega (\xi )}\right)\qquad {\mbox{and}}\qquad \beta (\xi ):=\frac{1}{2}\left(\widehat{v_0}(\xi )-i\frac{\widehat{v_1}(\xi )}{\omega (\xi )}\right),\) 
we have that \(\alpha +\beta =\widehat{v_0}\)  and \(-i\omega (\alpha -\beta )=\widehat{v_1}\) .
As a result, if
\(u(t,x):=\int _{\mathbb {R}}\left\lbrace \alpha (\xi )e^{-i\left[\xi x+\omega (\xi )\,t\right]}+\beta (\xi )e^{-i\left[\xi x-\omega (\xi )\,t\right]}\right\rbrace \,d\xi \,,\) 
we see that
\( u(0,x)=\int _{\mathbb {R}}\left\lbrace \alpha (\xi )e^{-i\xi x}+\beta (\xi )e^{-i\xi x}\right\rbrace \,d\xi =\int _{\mathbb {R}}\widehat{v_0}(\xi )\,e^{-i\xi x}\,d\xi =v_0(x)\) 
and
\( u_t(0,x)=\int _{\mathbb {R}}\left\lbrace -i\alpha (\xi ) \omega (\xi ) e^{-i \xi x }+i \beta (\xi )\omega (\xi )e^{-i\xi x }\right\rbrace \,d\xi =\int _{\mathbb {R}}\widehat{v_1}(\xi )\,e^{-i\xi x}\,d\xi =v_1(x),\) 
hence (REF ) provides a solution of (REF ). We can also rewrite (REF ) in the form given by (REF ).
Additionally, we observe that our assumptions
\(v_0,\,v_1\in \mathcal {S}(\mathbb {R})\)  allow us to state that \(\widehat{v_0},\,\widehat{v_1}\in \mathcal {S}(\mathbb {R})\) .
Moreover, for every \(t\in \mathbb {R}\) ,
we have that
\(1-\cos t\le \min \left\lbrace \frac{t^2}{2},2\right\rbrace ,\) 
hence it follows from (REF ) that
\(\omega ^2(\xi )&=&\frac{2\kappa }{\rho \,\delta ^{2\alpha }}\,\int _{-1}^{1}\frac{1-\cos (\xi \delta z)}{|z|^{1+2\alpha }}dz\\&\le &\frac{2\kappa }{\rho \,\delta ^{2\alpha }}\,\int _{-1}^{1}\frac{\min \left\lbrace \frac{\xi ^2\delta ^2 z^2}{2},2\right\rbrace }{|z|^{1+2\alpha }}dz\\&\le &\frac{2\kappa }{\rho \,\delta ^{2\alpha }}\,\left[\int _{\left\lbrace  |z|\le \min \left\lbrace \frac{2}{|\xi |\delta },1\right\rbrace \right\rbrace }{\frac{\xi ^2\delta ^2 z^2}{2|z|^{1+2\alpha }}}dz+\int _{\left\lbrace  \frac{2}{|\xi |\delta }<|z|\le 1\right\rbrace }\frac{2}{|z|^{1+2\alpha }}dz\right]\\&=&\frac{2\kappa }{\rho \,\delta ^{2\alpha }}\,\left[\frac{|\xi |^{2}\delta ^{2}}{2-2\alpha }\,\min \left\lbrace \frac{2^{2-2\alpha }}{|\xi |^{2-2\alpha }\delta ^{2-2\alpha }},1\right\rbrace +\frac{2}{\alpha }\chi _{\left( 0,1\right)}\left(\frac{2}{|\xi |\delta }\right) \,\left( \frac{|\xi |^{2\alpha }\delta ^{2\alpha }}{2^{2\alpha }}-1\right)\right],\) 
that gives (REF ).
We point out that
problem (REF )
reduces to the classical wave equation as \(\alpha \rightarrow 1^-\) ,
in a sense which is made precise in Lemma REF .
Similarly,
the explicit solution provided
in (REF )
and (REF ) approaches the one obtained
by Fourier methods for the classical wave equation,
as specified in Lemma REF .
Dispersion relation
We now deepen our analysis
of the dispersion relation introduced in (REF )
by supporting the estimate in (REF )
with some precise asymptotics:
Theorem 2.1 
For \(\delta >0\)  and \(0<\alpha <1\) , we have that
\(\lim _{\xi \rightarrow 0}\xi ^{-2}\omega ^2(\xi )= \frac{\kappa \,\delta ^{2(1-\alpha )}}{(1-\alpha )\,\rho }\) 
and
\(\lim _{\xi \rightarrow \pm \infty }|\xi |^{-2\alpha }\omega ^2(\xi )=\frac{4\kappa }{\rho }\,\int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau .\) 

We observe that
the asymptotics in (REF )
and (REF ) show a different power law behavior of the dispersion relation \(\omega \) 
at zero and at infinity. This different behavior is also confirmed numerically in Figure REF ,
where \(\omega \)  is plotted in logarithmic scale: as usual, in this setting, the two different power laws
correspond to straight lines with different slopes.
Furthermore, we point out that a suitable constitutive restriction on the elastic material parameter \(\kappa \) 
should take into account the asymptotic scaling stated in (REF ), namely
\(\kappa \sim \frac{1}{\delta ^{2(1-\alpha )}}.\) 
<FIGURE>Let \(\xi _j\)  be an infinitesimal sequence and
\( F_j(z):=\frac{1-\cos (\xi _j\delta z)}{\xi _j^2\,|z|^{1+2\alpha }}=\frac{\delta ^2 \,|z|^{1-2\alpha }\big (1-\cos (\xi _j\delta z)\big )}{(\xi _j \delta z)^2}.\) 
We point out that, for each \(z\in [-1,1]\) ,
\(\lim _{j\rightarrow +\infty } F_j(z)=\frac{\delta ^2 \,|z|^{1-2\alpha }}{2}.\) 
Additionally, recalling (REF ),
\( |F_j(z)|\le \frac{\xi _j^2\delta ^2 z^2}{2\xi _j^2\,|z|^{1+2\alpha }}=\frac{\delta ^2 |z|^{1-2\alpha }}{2}=:G(z).\) 
Since \(G\in L^1([-1,1])\) , we can use the Dominated Convergence Theorem and infer that
\( \lim _{j\rightarrow +\infty }\int _{-1}^1 \frac{1-\cos (\xi _j\delta z)}{\xi _j^2\,|z|^{1+2\alpha }}\,dz=\lim _{j\rightarrow +\infty }\int _{-1}^1 F_j(z)\,dz=\int _{-1}^1 \frac{\delta ^2 \,|z|^{1-2\alpha }}{2}\,dz=\frac{\delta ^2}{2-2\alpha }.\) 
From this equation and the definition of \(\omega \)  in (REF )
we plainly obtain the desired result in (REF ).
Moreover, using the substitution \(\tau :=|\xi |\delta z\) ,
\(\lim _{\xi \rightarrow \pm \infty }|\xi |^{-2\alpha }\omega ^2(\xi )&=&\lim _{\xi \rightarrow \pm \infty }\frac{4\kappa \,|\xi |^{-2\alpha }}{\rho \,\delta ^{2\alpha }}\,\int _{0}^{1}\frac{1-\cos (\xi \delta z)}{z^{1+2\alpha }}dz\\&=&\lim _{\xi \rightarrow \pm \infty }\frac{4\kappa }{\rho }\,\int _{0}^{|\xi |\delta }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau ,\) 
from which (REF ) plainly follows.
Now we present a sharpening of Theorem REF  in a logarithmic scale,
in view of an asymptotic as \(\alpha \rightarrow 1^-\) .
Theorem 2.2 
Let \(\delta >0\)  and \(\frac{1}{2}\le \alpha <1\) . Then, given \(b>0\)  there exists \(C>0\) , that depends only on \(b\)  and \(\delta \) ,
such that for all \(\xi \in \mathbb {R}\setminus (-b,b)\)
\( \left|\frac{\sqrt{(1-\alpha )\rho }\,\omega (\xi )}{\sqrt{\kappa }\,|\xi |^{\alpha }}-1\right|\le \sqrt{C\,(1-\alpha )}. \) 
Also, there exists \(c\in \left(0,\frac{1}{2}\right)\) , that depends only on \(b\) , \(\delta \)  and \(\kappa \) , such that
if \(1-\alpha \le c\)  then, for all \(\xi \in \mathbb {R}\setminus (-b,b)\) ,
\( \left|\log \frac{\sqrt{(1-\alpha )\rho }\,\omega (\xi )}{\sqrt{\kappa }}-\alpha \log |\xi |\right|\le \sqrt{C\,(1-\alpha )} .\) 

We observe that (REF ) gives a convergence of the dispersion relation
to a straight line in logarithmic scale (with an explicit error bound). Also (REF ) states that
this convergence is uniformWe cannot expect uniform convergence up to the origin. Indeed,
as we will see in (REF ), near the origin
\( \omega (\xi )=\frac{\sqrt{\kappa }|\xi |\delta ^{1-\alpha }}{\sqrt{(1-\alpha )\rho }}\left(1+O(\xi ^2)\right)\) 
and therefore
\(\left|\log \frac{\sqrt{(1-\alpha )\rho }\,\omega (\xi )}{\sqrt{\kappa }}-\alpha \log |\xi |\right|=(1-\alpha )\left|\log \delta +\log |\xi |\right|\) 
which diverges as \(\xi \rightarrow 0\) .
 outside the origin. For a numerical evidence of the
convergence of the dispersion relation
to a straight line in logarithmic scale see Figure REF .
<FIGURE>Let \(a_0\in (0,1)\) .
First of all, we claim that there exists \(C>0\)  depending only on \(a_0\)  such that for every \(t\ge a_0\)
\(\left|\int _0^t\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau -\frac{1}{4(1-\alpha )}\right|\le C.\) 
To check this, we distinguish two cases, according to whether \(t\in [a_0,1]\) 
or \(t>1\) . If \(t\in [a_0,1]\) , we use that
\( 1-\cos \tau -\frac{\tau ^2}{2}+\frac{\tau ^3}{6}\ge 0\qquad {\mbox{for all }}\tau \in \mathbb {R}\) 
and we see that
\(\int _0^t\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau \ge \int _0^t\frac{\frac{\tau ^2}{2}-\frac{\tau ^3}{6}}{\tau ^{1+2\alpha }}d\tau =\frac{t^{2 - 2 \alpha }}{4 (1-\alpha )}+\frac{t^{3 - 2 \alpha }}{6 (2 \alpha - 3)} .\) 
Similarly, since
\( 1-\cos \tau -\frac{\tau ^2}{2}\le 0\qquad {\mbox{for all }}\tau \in \mathbb {R},\) 
we obtain that
\(\int _0^t\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau \le \int _0^t\frac{\frac{\tau ^2}{2}}{\tau ^{1+2\alpha }}d\tau =\frac{t^{2 - 2 \alpha }}{4 (1-\alpha )} .\) 
By combining this and (REF ), we obtain that, for all \(t\in [a_0,1]\) ,
\(\left|\int _0^t\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau -\frac{1}{4 (1-\alpha )}\right|\le \frac{1-t^{2 - 2 \alpha }}{4 (1-\alpha )}+\frac{t^{3 - 2 \alpha }}{6 (3-2 \alpha )}\le \frac{1-a_0^{2(1 - \alpha )}}{4 (1-\alpha )}+\frac{1}{6}.\) 
Thus, since
\( a_0^{2(1 - \alpha )}=\exp \left((1-\alpha )\log (a_0^2)\right)\ge 1+(1-\alpha )\log (a_0^2),\) 
thanks to the convexity of the exponential function,
we conclude that
\(\left|\int _0^t\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau -\frac{1}{4 (1-\alpha )}\right|\le -\frac{\log (a_0^2)}{4 (1-\alpha )}+\frac{1}{6}.\) 
This proves (REF ) when \(t\in [a_0,1]\) . If instead \(t>1\) ,
we use (REF ) with \(t:=1\)  to see that
\(\left|\int _0^t\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau -\frac{1}{4 (1-\alpha )}\right|\le \left|\int _0^1\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau -\frac{1}{4 (1-\alpha )}\right|+\int _1^t\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau \le C+2\int _1^{+\infty }\frac{d\tau }{\tau ^{2}},\) 
from which we obtain (REF ) in this case as well.
Hence, combining (REF ) and (REF ), for all \(\xi \ge a_0/\delta \) ,
\(\frac{\sqrt{\kappa }\xi ^{\alpha }}{\sqrt{(1-\alpha )\rho }}\left|\omega (\xi )-\frac{\sqrt{\kappa }\xi ^{\alpha }}{\sqrt{(1-\alpha )\rho }}\right|&\le &\left(\omega (\xi )+\frac{\sqrt{\kappa }\xi ^{\alpha }}{\sqrt{(1-\alpha )\rho }}\right)\left|\omega (\xi )-\frac{\sqrt{\kappa }\xi ^{\alpha }}{\sqrt{(1-\alpha )\rho }}\right|\\&=&\left|\omega ^2(\xi )-\frac{\kappa \xi ^{2\alpha }}{(1-\alpha )\rho }\right|\\&=&\left|\frac{4\kappa }{\rho \,\delta ^{2\alpha }}\,\int _{0}^{1}\frac{1-\cos (\xi \delta z)}{z^{1+2\alpha }}dz - \frac{\kappa \xi ^{2\alpha }}{(1-\alpha )\rho }\right|\\&=&\frac{4\kappa \xi ^{2\alpha }}{\rho }\,\left|\int _{0}^{\xi \delta }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau - \frac{1}{4(1-\alpha )}\right|\\&\le &\frac{C\kappa \xi ^{2\alpha }}{\rho }\) 
and therefore the claim in (REF ) plainly follows by taking \(a_0:=b\delta \)  and recalling that \(\omega \) 
is an even function.
Moreover, we claim that
\(|\log (1+r)|\le 4|r|\qquad {\mbox{for every }}r\in \left[-\frac{1}{2},\frac{1}{2}\right].\) 
Indeed, suppose not, namely
\( \min _{\left[-\frac{1}{2},\frac{1}{2}\right]}\psi <0,\) 
where \(\psi (r):=4|r|-|\log (1+r)|\) . Let \(r_0\)  be the point attaining the above minimum.
Thus, since \(\psi (0)=0\) ,
\(\psi \left(-\frac{1}{2}\right)=2-|\log \frac{1}{2}|>0\) 
and
\(\psi \left(\frac{1}{2}\right)=2-\log \frac{3}{2}>0\) , necessarily \(\psi ^{\prime }(r_0)=0\)  and \(r_0\ne 0\) . This gives that
\( 0=\psi ^{\prime }(r_0)={\left\lbrace \begin{array}{ll}4-\frac{1}{1+r_0} &{\mbox{ if $r_0\in \left(0,\frac{1}{2}\right)$,}}\\-4+\frac{1}{1+r_0}&{\mbox{ if $r_0\in \left(-\frac{1}{2},0\right)$.}}\end{array}\right.}\) 
The above cases readily produce a contradiction, hence (REF )
is proved.
Using together (REF ) and (REF ) with \(r:=\frac{\sqrt{(1-\alpha )\rho }\,\omega (\xi )}{\sqrt{\kappa }\,|\xi |^{\alpha }}-1\) ,
we obtain that
\(\left|\log \frac{\sqrt{(1-\alpha )\rho }\,\omega (\xi )}{\sqrt{\kappa }}-\alpha \log |\xi |\right|=\left|\log \frac{\sqrt{(1-\alpha )\rho }\,\omega (\xi )}{\sqrt{\kappa }\,|\xi |^{\alpha }}\right|\le 4\left|\frac{\sqrt{(1-\alpha )\rho }\,\omega (\xi )}{\sqrt{\kappa }\,|\xi |^{\alpha }}-1\right|\le 4\sqrt{C\,(1-\alpha )}\) 
as long as \(\sqrt{C\,(1-\alpha )}\le \frac{1}{2}\) .
This establishes (REF ), up to renaming \(C\) .
As it is well known, in wave propagation in dispersive medium, initiated by Lord Rayleigh ([2]}), a crucial role is played by the notion of group velocity which is given by the derivative of the dispersion with respect to the frequency variable. Indeed, this remarkable role stays in the property that the group velocity corresponds to the velocity of energy transport ([3]}) in a large class of so called nondissipative media.
Therefore,
now we are going to extend the asymptotics of
Theorem REF  to the derivatives of the dispersion, to the aim of obtaining quantitative estimates on the behavior of the group velocity.
Theorem 2.3 
For \(\delta >0\)  and \(0\le \alpha <1\) ,
we have that
\(&&\lim _{\xi \rightarrow 0^\pm } \omega ^{\prime }(\xi )=\pm \frac{\sqrt{\kappa }\delta ^{1-\alpha }}{\sqrt{(1-\alpha )\rho } },\\&&\lim _{\xi \rightarrow \pm \infty }|\xi |^{1-\alpha }\omega ^{\prime }(\xi )=\pm 2\alpha \,\sqrt{\frac{\kappa }{\rho }\,\int _{0}^{+\infty }\frac{1-\cos \tau }{ \tau ^{1+2\alpha }}d\tau },\\&&\lim _{\xi \rightarrow 0^\pm }|\xi |^{-1}\omega ^{\prime \prime }(\xi )=-\frac{\sqrt{\kappa (1-\alpha )}\;\delta ^{3-\alpha } }{4(2-\alpha )\sqrt{\rho }},\\&&\lim _{\xi \rightarrow 0^\pm }\omega ^{\prime \prime }(\xi )=0,\\&&{\mbox{ if }}\alpha \in \left(\frac{1}{2},1\right){\mbox{ then }}\quad \lim _{\xi \rightarrow \pm \infty }|\xi |^{2-\alpha } \omega ^{\prime \prime }(\xi )=-\frac{2\alpha (1-\alpha )\sqrt{\kappa }}{\sqrt{\rho }}\left(\int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}\,d\tau \right)^{1/2},\\&&{\mbox{ if }}\alpha \in \left[0,\frac{1}{2}\right){\mbox{ then }}\quad \liminf _{\xi \rightarrow \pm \infty }|\xi |^{1+\alpha } \omega ^{\prime \prime }(\xi )=-\frac{\sqrt{\kappa }\delta ^{1-2\alpha }}{\sqrt{\rho }}\left(\int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau \right)^{-1/2}\nonumber \\&&\qquad \qquad \qquad \qquad \qquad \qquad <\frac{\sqrt{\kappa }\delta ^{1-2\alpha }}{\sqrt{\rho }}\left(\int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau \right)^{-1/2}\nonumber \\&&\qquad \qquad \qquad \qquad \qquad \qquad =\limsup _{\xi \rightarrow \pm \infty }|\xi |^{1+\alpha } \omega ^{\prime \prime }(\xi ),\\&&{\mbox{ if }}\alpha = \frac{1}{2}{\mbox{ then }}\quad \liminf _{\xi \rightarrow \pm \infty }|\xi |^{3/2} \omega ^{\prime \prime }(\xi )\nonumber \\&&\qquad \qquad \qquad \qquad =\left[-\frac{\sqrt{\kappa }}{\sqrt{\rho }}\left({ \int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{2}}d\tau }\right)^{-1/2}-\frac{1}{2}\left(\frac{ \kappa }{ \rho }{ \int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{2}}d\tau }\right)^{1/2}\right]\nonumber \\&&\qquad \qquad \qquad \qquad <\left[\frac{\sqrt{\kappa }}{ \sqrt{\rho }}\left({ \int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{2}}d\tau }\right)^{-1/2}-\frac{1}{2}\left(\frac{ \kappa }{ \rho }{ \int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{2}}d\tau }\right)^{1/2}\right]\nonumber \\&&\qquad \qquad \qquad \qquad =\limsup _{\xi \rightarrow \pm \infty }|\xi |^{3/2} \omega ^{\prime \prime }(\xi ).\) 

We stress that Theorem REF  highlights a number of special features of the dispersion relation.
Indeed, it follows from the asymptotics in (REF ) and () of
Theorem REF 
that \(\omega ^{\prime }\)  has a jump discontinuity at the origin (hence \(\omega \)  presents a corner), but \(\omega ^{\prime \prime }\)  (as a function defined in \(\mathbb {R}\setminus \lbrace 0\rbrace \) )
can be extended continuously through the origin.
<FIGURE>Moreover, the convexity properties of the dispersion relation
present an interesting dependence on \(\alpha \) . Specifically,
when \(\alpha \in \left(\frac{1}{2},1\right)\) , formula () in
Theorem REF 
gives that \(\omega ^{\prime \prime }\)  is negative at infinity and therefore \(\omega \)  is concave at infinity.
Instead, when \(\alpha \in \left(0,\frac{1}{2}\right]\) ,
the asymptotics in formulas () and () of
Theorem REF  state that \(\omega ^{\prime \prime }\)  changes
sign infinitely many times at infinity and consequently
in this range the dispersion relation \(\omega \)  switches from convex to concave infinitely often. Besides detailed
analytic proofs, we also provide
numerical confirmations of these phenomena.
In particular, the function \(\omega ^{\prime \prime }\)  is plotted in Figure REF :
notice that \(\omega ^{\prime \prime }\)  is shown to intersect the horizontal axis infinitely many times when \(\alpha \in \left(0,\frac{1}{2}\right]\) 
in agreement with () and () (and differently from the case \(\alpha \in \left(\frac{1}{2},1\right)\) 
which instead is in agreement with ()).
<FIGURE>An additional interesting feature showcased by Theorem REF 
is that the derivatives of the the dispersion relation do not inherit the “natural decay at infinity”
from the original function. In particular, while \(\omega \)  at infinity behaves like \(|\xi |^\alpha \) 
in light of (REF ), contrary to the usual situations it is not always true in this setting that \(\omega ^{\prime \prime }\)  behaves at infinity
like \(|\xi |^{\alpha -2}\)  (that is, like \(\frac{\omega }{\xi ^2}\) ). More precisely,
while this is true when \(\alpha \in \left(\frac{1}{2},1\right)\) ,
thanks to formula () in
Theorem REF , in the range \(\alpha \in \left(0,\frac{1}{2}\right]\)  the behavior is completely different
and the leading order happens to be \(|\xi |^{1+\alpha }\)  (surprisingly corresponding to \(\frac{\omega }{\xi ^{-1}}\) ,
and also presenting oscillatory behaviors).
The different power law behaviors of \(\omega ^{\prime \prime }\)  at infinity stated in (), () and ()
are also numerically confirmed by Figures REF  and REF .
In particular, Figure REF  showcases a numerical plot of \(|\xi |^{2-\alpha }\omega ^{\prime \prime }\) 
that confirms the convergence at infinity if \(\alpha \in \left(\frac{1}{2},1\right)\) ,
in agreement with (), its divergence if \(\alpha \in \left(0,\frac{1}{2}\right)\) ,
in agreement with (), its oscillatory boundedness \(\alpha =\frac{1}{2}\) ,
in agreement with (). Instead,
Figure REF  showcases a numerical plot of \(|\xi |^{1+\alpha }\omega ^{\prime \prime }\) 
that confirms the divergence at infinity if \(\alpha \in \left(\frac{1}{2},1\right)\) ,
in agreement with (), and its bounded oscillatory behavior
if \(\alpha \in \left(0,\frac{1}{2}\right]\) ,
in agreement with () and ().
We also notice that when \(\alpha =\frac{1}{2}\)  the plots in Figures REF 
and REF  agree, consistently with the fact that \(2-\alpha =1+\alpha \)  in this specific case.
<FIGURE>The unusual phenomena detected
in (),
() and ()
are deeply related to the nonlocal nature of the problem
and to the appearance of divergent singular integrals in the formal expansions of the dispersion relation.
This important technical details prevent us to use lightly formal expansions
and soft arguments of general flavor since, roughly speaking, terms that are usually “negligible”
in a standard expansion may become “dominant” in our setting since they may end
up being multiplied
by a “divergent” coefficient induced by a singular integral and, quite interestingly,
as emphasized
by the asymptotics in formulas () and ()
of Theorem REF ,
these new significant terms
may even be of oscillatory type.
The “numerology” of Theorem REF  is also somewhat
interesting since all the coefficients appearing in the asymptotics
are determined explicitly (and finding an explicit representation
of a coefficient is often the most direct way to prove that
it is finite as well). As a matter of fact, though we do not make
use of this fact, we mention that
the trigonometric integral appearing in
Theorem REF  (as well as in (REF ))
can be computed in terms
of the Euler Gamma Function, since
\(\int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau =-\cos (\pi \alpha )\Gamma (-2\alpha ),\) 
with the left hand side
continuously extended to the value \(\frac{\pi }{2}\)  when \(\alpha =\frac{1}{2}\) ,
see Appendix 
for an elementary
(or Appendix  for a shorter,
but more sophisticated) proof of (REF ).
See also Figure REF  for a numerical confirmation of (REF ):
indeed, in Figure REF  the plots of the functions \(\alpha \mapsto \int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau \)  and \(\alpha \mapsto -\cos (\pi \alpha )\Gamma (-2\alpha )\) 
are given and one can observe that the two graphs are the same, up to a sign change,
in full agreement with (REF ).
<FIGURE>It is also interesting to recall that the threshold \(\alpha =\frac{1}{2}\) 
that emerges in
formulas (), () and () of
Theorem REF  is also an important threshold
for several other nonlocal problems, see
e.g. [4]}, [5]}, [6]}.
Differentiating (REF )
and using the change of variable \(w:=\xi \delta z\) 
we see that
\(\begin{split}&2\omega (\xi )\omega ^{\prime }(\xi )=\frac{d}{d\xi }\omega ^2(\xi )=\frac{d}{d\xi }\left(\frac{2\kappa }{\rho \,\delta ^{2\alpha }}\,\int _{-1}^{1}\frac{1-\cos (\xi \delta z)}{|z|^{1+2\alpha }}dz\right)=\frac{2\kappa }{\rho \,\delta ^{2\alpha -1}}\,\int _{-1}^{1}\frac{z\sin (\xi \delta z)}{|z|^{1+2\alpha }}dz\\&\qquad \qquad \qquad =\frac{4\kappa }{\rho \,\delta ^{2\alpha -1}}\,\int _{0}^{1}\frac{z\sin (\xi \delta z)}{|z|^{1+2\alpha }}dz=\frac{4\kappa |\xi |^{2\alpha -1}}{\rho }\,\int _{0}^{\xi \delta }\frac{w\sin w}{ |w|^{1+2\alpha }}dw\\&\qquad \qquad \qquad =\frac{4\kappa |\xi |^{2\alpha -1}}{\rho }\,\int _{0}^{\xi \delta }\frac{\sin |w|}{ |w|^{2\alpha }}dw\,.\end{split}\) 
From this and (REF ), using l'Hôpital's Rule we find that
\(&&\frac{2\sqrt{\kappa }\,\delta ^{1-\alpha }}{\sqrt{(1-\alpha )\,\rho }}\lim _{\xi \rightarrow 0^\pm } \omega ^{\prime }(\xi )=\lim _{\xi \rightarrow 0^\pm }\frac{2\omega (\xi )\omega ^{\prime }(\xi )}{|\xi |}=\lim _{\xi \rightarrow 0^\pm }\frac{4\kappa }{\rho |\xi |^{2-2\alpha }}\,\int _{0}^{\xi \delta }\frac{\sin |w|}{ |w|^{ 2\alpha }}dw\\&&\qquad =\lim _{\xi \rightarrow 0^\pm }\frac{4\kappa \delta }{(2-2\alpha )\rho |\xi |^{-2\alpha }\xi }\,\frac{ \sin |\xi \delta |}{ |\xi \delta |^{2\alpha }}=\lim _{\xi \rightarrow 0^\pm }\frac{2\kappa \delta ^{1-2\alpha }}{(1-\alpha )\rho }\,\frac{\sin |\xi \delta |}{ \xi }\\&&\qquad =\pm \frac{2\kappa \delta ^{2-2\alpha }}{(1-\alpha )\rho }.\) 
From this, we obtain formula (REF ) in Theorem REF .
Similarly, using (REF ) and (REF ),
\(&&\frac{4\sqrt{\kappa }}{\sqrt{\rho }}\,\sqrt{\int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau }\lim _{\xi \rightarrow \pm \infty }|\xi |^{1-\alpha }\omega ^{\prime }(\xi )=\lim _{\xi \rightarrow \pm \infty }2|\xi |^{1-2\alpha }\omega (\xi )\omega ^{\prime }(\xi )\\&&\qquad =\lim _{\xi \rightarrow \pm \infty }\frac{4\kappa }{\rho }\,\int _{0}^{\xi \delta }\frac{w\sin w}{ |w|^{1+2\alpha }}dw=\frac{4\kappa }{\rho }\,\int _{0}^{\pm \infty }\frac{\sin | w|}{ |w|^{2\alpha }}dw=\pm \frac{4\kappa }{\rho }\,\int _{0}^{+\infty }\frac{\sin \tau }{ \tau ^{2\alpha }}d\tau .\) 
We stress that when \(\alpha \in \left(0,\frac{1}{2}\right]\) 
the function \(\frac{\sin \tau }{ \tau ^{2\alpha }}\)  is not
Lebesgue summable in \((0,+\infty )\) ; nevertheless, for
every \(\alpha \in (0,1)\)  one can write the improper Riemann integral
\(\begin{split}&\int _{0}^{+\infty }\frac{\sin \tau }{ \tau ^{2\alpha }}d\tau :=\lim _{R\rightarrow +\infty }\int _{0}^{R}\frac{\sin \tau }{ \tau ^{2\alpha }}d\tau =\lim _{R\rightarrow +\infty }\int _{0}^{R}\left[\left(\frac{1-\cos \tau }{ \tau ^{2\alpha }}\right)^{\prime }+\frac{2\alpha (1-\cos \tau )}{ \tau ^{1+2\alpha }}\right]d\tau \\&\qquad =\lim _{R\rightarrow +\infty }\frac{1-\cos R}{R^{2\alpha }}-\lim _{t\rightarrow 0}\frac{1-\cos t}{t^{2\alpha }}+2\alpha \int _{0}^{+\infty }\frac{1-\cos \tau }{ \tau ^{1+2\alpha }}d\tau \\&\qquad =0-\lim _{t\rightarrow 0}t^{2-2\alpha }\,\frac{1-\cos t}{t^{2}}+2\alpha \int _{0}^{+\infty }\frac{1-\cos \tau }{ \tau ^{1+2\alpha }}d\tau =2\alpha \int _{0}^{+\infty }\frac{1-\cos \tau }{ \tau ^{1+2\alpha }}d\tau .\end{split}\) 
These observations lead to
\( \frac{4\sqrt{\kappa }}{\sqrt{\rho }}\,\sqrt{\int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau }\lim _{\xi \rightarrow \pm \infty }|\xi |^{1-\alpha }\omega ^{\prime }(\xi )=\pm \frac{8\alpha \kappa }{\rho }\,\int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau ,\) 
which produces formula () in Theorem REF .
Also, differentiating once more in (REF ), for all \(\xi \ne 0\) ,
\(\begin{split}&\omega (\xi )\omega ^{\prime \prime }(\xi )+(\omega ^{\prime }(\xi ))^2=\frac{d}{d\xi }\big (\omega (\xi )\omega ^{\prime }(\xi )\big )=\frac{d}{d\xi }\left(\frac{2\kappa }{\rho \,\delta ^{2\alpha -1}}\,\int _{0}^{1}\frac{\sin (\xi \delta z)}{z^{2\alpha }}dz\right)\\&\qquad =\frac{2\kappa \delta ^{2-2\alpha }}{\rho }\,\int _{0}^{1}\frac{\cos (\xi \delta z)}{z^{2\alpha -1}}dz\,.\end{split}\) 
Thus, using again (REF ),
\(\begin{split}\omega (\xi )\omega ^{\prime \prime }(\xi )\,&=\,\frac{2\kappa \delta ^{2-2\alpha }}{\rho }\,\int _{0}^{1}\frac{\cos (\xi \delta z)}{z^{2\alpha -1}}dz-(\omega ^{\prime }(\xi ))^2\\&=\,\frac{2\kappa \delta ^{2-2\alpha }}{\rho }\,\int _{0}^{1}\frac{\cos (\xi \delta z)}{z^{2\alpha -1}}dz-\left(\frac{2\kappa }{\rho \,\delta ^{2\alpha -1} \omega (\xi )}\,\int _{0}^{1}\frac{\sin (\xi \delta z)}{z^{2\alpha }}dz\right)^2.\end{split}\) 
Also, in light of (REF ), as \(\xi \rightarrow 0^+\) ,
\(\begin{split}&\omega ^2(\xi )=\frac{2\kappa }{\rho \,\delta ^{2\alpha }}\,\int _{-1}^{1}\frac{1-\cos (\xi \delta z)}{|z|^{1+2\alpha }}dz=\frac{4\kappa }{\rho \,\delta ^{2\alpha }}\,\int _{0}^1\frac{\frac{(\xi \delta z)^2}{2}-\frac{(\xi \delta z)^4}{24}+O(\xi ^6z^6)}{z^{1+2\alpha }}dz\\&\qquad =\frac{4\kappa }{\rho \,\delta ^{2\alpha }}\left(\frac{\xi ^2\delta ^2}{4(1-\alpha )}-\frac{\xi ^4\delta ^4}{48(2-\alpha )}+O(\xi ^6)\right)=\frac{\kappa \xi ^2\delta ^{2(1-\alpha )}}{(1-\alpha )\rho }-\frac{\kappa \xi ^4\delta ^{2(2-\alpha )}}{12(2-\alpha )\rho }+O(\xi ^6)\\&\qquad =\frac{\kappa \xi ^2\delta ^{2(1-\alpha )}}{(1-\alpha )\rho }\left(1-\frac{(1-\alpha )\xi ^2\delta ^{2}}{12(2-\alpha )}+O(\xi ^4)\right),\end{split}\) 
which can be seen as an enhanced version of (REF ).
As a result,
\(\left(\frac{2\kappa }{\rho \,\delta ^{2\alpha -1} \omega (\xi )}\right)^2=\frac{4(1-\alpha )\kappa }{ \xi ^2\delta ^{2 \alpha }\rho \left(1-\frac{(1-\alpha )\xi ^2\delta ^{2}}{12(2-\alpha )}+O(\xi ^4)\right)}=\frac{4(1-\alpha )\kappa }{ \xi ^2\delta ^{2 \alpha }\rho }\left(1+\frac{(1-\alpha )\xi ^2\delta ^{2}}{12(2-\alpha )}+O(\xi ^4)\right).\) 
Hence, by (REF ), as \(\xi \rightarrow 0^+\) ,
\(&&\frac{\sqrt{\kappa }\xi \delta ^{1-\alpha }}{\sqrt{(1-\alpha )\rho }}\sqrt{1+O(\xi ^2) }\;\omega ^{\prime \prime }(\xi )\\&=&\frac{2\kappa \delta ^{2-2\alpha }}{\rho }\,\int _{0}^{1}\frac{\cos (\xi \delta z)}{z^{2\alpha -1}}dz-\frac{4(1-\alpha )\kappa }{ \xi ^2\delta ^{2 \alpha }\rho }\left(1+\frac{(1-\alpha )\xi ^2\delta ^{2}}{12(2-\alpha )}+O(\xi ^4)\right)\left(\int _{0}^{1}\frac{\sin (\xi \delta z)}{z^{2\alpha }}dz\right)^2.\) 
Thus, noticing that, as \(\xi \rightarrow 0^+\) ,
\(&& \int _{0}^{1}\frac{\cos (\xi \delta z)}{z^{2\alpha -1}}dz=\int _{0}^{1}\frac{1-\frac{(\xi \delta z)^2}{2}+O(\xi ^4z^4)}{z^{2\alpha -1}}dz=\frac{1}{2(1-\alpha )}-\frac{\xi ^2\delta ^2}{4(2-\alpha )}+O(\xi ^4)\) 
and
\(&&\int _{0}^{1}\frac{\sin (\xi \delta z)}{z^{2\alpha }}dz=\int _{0}^{1}\frac{\xi \delta z-\frac{(\xi \delta z)^3}{6}+O(\xi ^5 z^5)}{z^{2\alpha }}dz=\frac{\xi \delta }{2(1-\alpha )}-\frac{\xi ^3\delta ^3}{12(2-\alpha )}+O(\xi ^5),\) 
we conclude that
\(&&\frac{\sqrt{\kappa }\xi \delta ^{1-\alpha }}{\sqrt{(1-\alpha )\rho }}\sqrt{1+O(\xi ^2) }\;\omega ^{\prime \prime }(\xi )\\&=&\frac{2\kappa \delta ^{2-2\alpha }}{\rho }\left(\frac{1}{2(1-\alpha )}-\frac{\xi ^2\delta ^2}{4(2-\alpha )}+O(\xi ^4)\right)\\&&\quad -\frac{4(1-\alpha )\kappa \delta ^{2-2 \alpha }}{ \rho }\left(1+\frac{(1-\alpha )\xi ^2\delta ^{2}}{12(2-\alpha )}+O(\xi ^4)\right)\left(\frac{1}{2(1-\alpha )}-\frac{\xi ^2\delta ^2}{12(2-\alpha )}+O(\xi ^4)\right)^2\\&=&\frac{2\kappa \delta ^{2-2\alpha }}{\rho }\left(\frac{1}{2(1-\alpha )}-\frac{\xi ^2\delta ^2}{4(2-\alpha )}+O(\xi ^4)\right)\\&&\quad -\frac{4(1-\alpha )\kappa \delta ^{2-2 \alpha }}{ \rho }\left(1+\frac{(1-\alpha )\xi ^2\delta ^{2}}{12(2-\alpha )}+O(\xi ^4)\right)\left(\frac{1}{4(1-\alpha )^2}-\frac{\xi ^2\delta ^2}{12(1-\alpha )(2-\alpha )}+O(\xi ^4)\right)\\&=&\frac{\kappa \delta ^{2-2\alpha }}{\rho }\left(\frac{1}{ 1-\alpha }-\frac{\xi ^2\delta ^2}{2(2-\alpha )}+O(\xi ^4)\right)-\frac{\kappa \delta ^{2-2 \alpha }}{ \rho }\left(\frac{1}{ 1-\alpha }+\frac{\xi ^2\delta ^{2}}{12(2-\alpha )}-\frac{\xi ^2\delta ^2}{3(2-\alpha )}+O(\xi ^4)\right)\\&=&\frac{\kappa \delta ^{2-2\alpha }}{\rho }\left( -\frac{\xi ^2\delta ^2}{2(2-\alpha )}-\frac{\xi ^2\delta ^{2}}{12(2-\alpha )}+\frac{\xi ^2\delta ^2}{3(2-\alpha )}+O(\xi ^4)\right)\\&=&-\frac{\kappa \delta ^{4-2\alpha }\xi ^2}{4(2-\alpha )\rho }+O(\xi ^4).\) 
This entails that
\(\frac{\sqrt{\kappa }\delta ^{1-\alpha }}{\sqrt{(1-\alpha )\rho }}\,\lim _{\xi \rightarrow 0^+}\xi ^{-1}\omega ^{\prime \prime }(\xi )=-\frac{\kappa \delta ^{4-2\alpha } }{4(2-\alpha )\rho }\) 
and therefore
\(\lim _{\xi \rightarrow 0^+}|\xi |^{-1}\omega ^{\prime \prime }(\xi )=-\frac{\sqrt{\kappa (1-\alpha )}\;\delta ^{3-\alpha } }{4(2-\alpha )\sqrt{\rho }}.\) 
Since \(\omega \)  (and thus \(\omega ^{\prime \prime }\) ) is an even function, we also have that
\(\lim _{\xi \rightarrow 0^-}|\xi |^{-1}\omega ^{\prime \prime }(\xi )=-\frac{\sqrt{\kappa (1-\alpha )}\;\delta ^{3-\alpha } }{4(2-\alpha )\sqrt{\rho }}.\) 
This and (REF ) give formula () in Theorem REF 
(from which formula ()
in Theorem REF  follows at once).
Let now \(\xi >0\) . From (REF ),
\(\omega (\xi )=\left(\frac{4\kappa }{\rho \,\delta ^{2\alpha }}\,\int _{0}^{1}\frac{1-\cos (\xi \delta z)}{z^{1+2\alpha }}dz\right)^{1/2}=\xi ^\alpha \left(\frac{4\kappa }{\rho }\,\int _{0}^{\xi \delta }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau \right)^{1/2}\) 
and thereforeWe observe that an alternative
proof of formula () in Theorem REF 
follows from (REF ), namely
\(\lim _{\xi \rightarrow +\infty }\xi ^{1-\alpha }\omega ^{\prime }(\xi )&=&\lim _{\xi \rightarrow +\infty }\left[\alpha \left(\frac{4\kappa }{\rho }\,\int _{0}^{\xi \delta }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau \right)^{1/2}+\frac{2\kappa \xi ^{-2\alpha }(1-\cos (\xi \delta ))}{\rho \delta ^{2\alpha }}\left(\frac{4\kappa }{\rho }\,\int _{0}^{\xi \delta }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau \right)^{-1/2}\right]\\&=&\alpha \left(\frac{4\kappa }{\rho }\,\int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau \right)^{1/2}+0,\) 
and formula () of Theorem REF  follows from this and
the fact that \(\omega ^{\prime }\)  is odd.

\(\omega ^{\prime }(\xi )=\alpha \xi ^{\alpha -1}\left(\frac{4\kappa }{\rho }\,\int _{0}^{\xi \delta }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau \right)^{1/2}+\frac{2\kappa \xi ^{-\alpha -1}(1-\cos (\xi \delta ))}{\rho \delta ^{2\alpha }}\left(\frac{4\kappa }{\rho }\,\int _{0}^{\xi \delta }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau \right)^{-1/2}.\) 
Taking
one more derivative,
\(\begin{split}&\omega ^{\prime \prime }(\xi )\,=\,\alpha (\alpha -1)\xi ^{\alpha -2}\left(\frac{4\kappa }{\rho }\,\int _{0}^{\xi \delta }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau \right)^{1/2}\\&\qquad \qquad +\frac{2\alpha \kappa \xi ^{-\alpha -2}(1-\cos (\xi \delta ))}{\rho \delta ^{2\alpha }}\left(\frac{4\kappa }{\rho }\,\int _{0}^{\xi \delta }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau \right)^{-1/2}\\&\qquad \qquad -\frac{2(\alpha +1)\kappa \xi ^{-\alpha -2}(1-\cos (\xi \delta ))}{\rho \delta ^{2\alpha }}\left(\frac{4\kappa }{\rho }\,\int _{0}^{\xi \delta }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau \right)^{-1/2}\\&\qquad \qquad +\frac{2\kappa \xi ^{-\alpha -1}\sin (\xi \delta )}{\rho \delta ^{2\alpha -1}}\left(\frac{4\kappa }{\rho }\,\int _{0}^{\xi \delta }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau \right)^{-1/2}\\&\qquad \qquad -\frac{4\kappa ^2\xi ^{-3\alpha -2}(1-\cos (\xi \delta ))^2}{\rho ^2\delta ^{4\alpha }}\left(\frac{4\kappa }{\rho }\,\int _{0}^{\xi \delta }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau \right)^{-3/2}.\end{split}\) 
That is, as \(\xi \rightarrow +\infty \) ,
\(\begin{split}&\omega ^{\prime \prime }(\xi )\,=\,\alpha (\alpha -1)\xi ^{\alpha -2}\left(\frac{4\kappa }{\rho }\,\int _{0}^{\xi \delta }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau \right)^{1/2}\\&\qquad \qquad +\frac{2\kappa \xi ^{-\alpha -1}\sin (\xi \delta )}{\rho \delta ^{2\alpha -1}}\left(\frac{4\kappa }{\rho }\,\int _{0}^{\xi \delta }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau \right)^{-1/2}+o(\xi ^{\alpha -2})+o(\xi ^{ -\alpha -1}).\end{split}\) 
Consequently, if \(\alpha \in \left(\frac{1}{2},1\right)\) ,
\(|\xi |^{2-\alpha } \omega ^{\prime \prime }(\xi )=\alpha (\alpha -1)\left(\frac{4\kappa }{\rho }\,\int _{0}^{\xi \delta }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau \right)^{1/2}+o(1)\) 
as \(\xi \rightarrow +\infty \) .
Similarly, if \(\alpha \in \left(0,\frac{1}{2}\right)\) ,
one deduces from (REF ) that
\(|\xi |^{1+\alpha } \omega ^{\prime \prime }(\xi )=\frac{2\kappa \sin (\xi \delta )}{\rho \delta ^{2\alpha -1}}\left(\frac{4\kappa }{\rho }\,\int _{0}^{\xi \delta }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau \right)^{-1/2}+o(1)\) 
as \(\xi \rightarrow +\infty \) .
And also, if \(\alpha =\frac{1}{2}\) , we get from (REF )
that
\(|\xi |^{3/2} \omega ^{\prime \prime }(\xi )=-\frac{1}{4}\left(\frac{4\kappa }{\rho }\,\int _{0}^{\xi \delta }\frac{1-\cos \tau }{\tau ^{2}}d\tau \right)^{1/2}+\frac{2\kappa \sin (\xi \delta )}{\rho }\left(\frac{4\kappa }{\rho }\,\int _{0}^{\xi \delta }\frac{1-\cos \tau }{\tau ^{2}}d\tau \right)^{-1/2}+o(1)\) 
as \(\xi \rightarrow +\infty \) .
These observations (and the fact that \(\omega ^{\prime \prime }\)  is an even function)
lead to formulas (), () and ()
of Theorem REF .
For completeness, we now provide alternative proofs for
the statements in (), () and ()
in Theorem REF .
To this end,
using (REF ), we observe that,
when \(\alpha \in \left[\frac{1}{2},1\right)\) ,
\(&&\liminf _{R\rightarrow +\infty }\int _{0}^{R}\frac{\cos \tau }{\tau ^{2\alpha -1}}d\tau =\liminf _{R\rightarrow +\infty }\int _{0}^{R}\left(\left(\frac{\sin \tau }{\tau ^{2\alpha -1}}\right)^{\prime }+(2\alpha -1)\frac{\sin \tau }{\tau ^{2\alpha }}\right)d\tau \\&&\qquad =\liminf _{R\rightarrow +\infty }\frac{\sin R}{R^{2\alpha -1}}+(2\alpha -1)\int _{0}^{+\infty }\frac{\sin \tau }{\tau ^{2\alpha }}d\tau \\&&\qquad ={\left\lbrace \begin{array}{ll}-1& {\mbox{ if }}\alpha =\frac{1}{2},\\\displaystyle 2\alpha (2\alpha -1)\int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}\,d\tau & {\mbox{ if }}\alpha \in \left(\frac{1}{2},1\right).\end{array}\right.}\) 
Similarly,
\(&&\limsup _{R\rightarrow +\infty }\int _{0}^{R}\frac{\cos \tau }{\tau ^{2\alpha -1}}d\tau ={\left\lbrace \begin{array}{ll}1& {\mbox{ if }}\alpha =\frac{1}{2},\\ \displaystyle 2\alpha (2\alpha -1)\int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}\,d\tau & {\mbox{ if }}\alpha \in \left(\frac{1}{2},1\right).\end{array}\right.}\) 
Accordingly, recalling (REF ) and (REF ), exploiting (REF )
once again, and using the substitution \(\tau :=|\xi |\delta z\) ,
if \(\alpha \in \left[\frac{1}{2},1\right)\) ,
\(&&\sqrt{ \frac{4\kappa }{\rho }\,\int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau }\liminf _{\xi \rightarrow \pm \infty }|\xi |^{2-\alpha } \omega ^{\prime \prime }(\xi )\\&=&\liminf _{\xi \rightarrow \pm \infty }|\xi |^{2-2\alpha }\omega (\xi )\omega ^{\prime \prime }(\xi )\\&=&\liminf _{\xi \rightarrow \pm \infty }|\xi |^{2-2\alpha }\left(\frac{2\kappa \delta ^{2-2\alpha }}{\rho }\,\int _{0}^{1}\frac{\cos (\xi \delta z)}{z^{2\alpha -1}}dz-\left(\frac{2\kappa }{\rho \,\delta ^{2\alpha -1} \omega (\xi )}\,\int _{0}^{1}\frac{\sin (\xi \delta z)}{z^{2\alpha }}dz\right)^2\right)\\&=&\liminf _{\xi \rightarrow \pm \infty }\frac{2\kappa }{\rho }\,\int _{0}^{|\xi |\delta }\frac{\cos \tau }{\tau ^{2\alpha -1}}d\tau -\left(\frac{2\kappa \,|\xi |^{\alpha }}{\rho \, \omega (\xi )}\,\int _{0}^{|\xi |\delta }\frac{\sin \tau }{\tau ^{2\alpha }}d\tau \right)^2\\&=&\frac{2\kappa }{\rho }\,\liminf _{R\rightarrow +\infty }\int _{0}^{R}\frac{\cos \tau }{\tau ^{2\alpha -1}}d\tau -\left(\frac{\kappa }{\rho \, \sqrt{ \frac{\kappa }{\rho }\,\int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau }}\,\int _{0}^{+\infty }\frac{\sin \tau }{\tau ^{2\alpha }}d\tau \right)^2\\&=&\frac{2\kappa }{\rho }\,\liminf _{R\rightarrow +\infty }\int _{0}^{R}\frac{\cos \tau }{\tau ^{2\alpha -1}}d\tau -\left(\frac{2\alpha \sqrt{\kappa }}{\sqrt{\rho }} \sqrt{ \int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau }\right)^2\\&=&{\left\lbrace \begin{array}{ll}\displaystyle -\frac{2\kappa }{\rho }+\frac{ \kappa }{ \rho }{ \int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{2}}d\tau }& {\mbox{ if }}\alpha =\frac{1}{2},\\\\\displaystyle \frac{4\alpha (\alpha -1)\kappa }{\rho }\int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}\,d\tau & {\mbox{ if }}\alpha \in \left(\frac{1}{2},1\right).\end{array}\right.}\) 
Similarly,
\(&&\sqrt{ \frac{4\kappa }{\rho }\,\int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau }\limsup _{\xi \rightarrow \pm \infty }|\xi |^{2-\alpha } \omega ^{\prime \prime }(\xi )={\left\lbrace \begin{array}{ll}\displaystyle \frac{2\kappa }{\rho }+\frac{ \kappa }{ \rho }{ \int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{2}}d\tau }& {\mbox{ if }}\alpha =\frac{1}{2},\\\\\displaystyle \frac{4\alpha (\alpha -1)\kappa }{\rho }\int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}\,d\tau & {\mbox{ if }}\alpha \in \left(\frac{1}{2},1\right).\end{array}\right.}\) 
These observation give formulas () and ()
in Theorem REF .
Besides, when \(\alpha \in \left(0,\frac{1}{2}\right)\) ,
we can infer from (REF ), (REF ) and (REF ),
that
\(&&\sqrt{ \frac{4\kappa }{\rho }\,\int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau }\liminf _{\xi \rightarrow \pm \infty }|\xi |^{1+\alpha } \omega ^{\prime \prime }(\xi )\\&=&\liminf _{\xi \rightarrow \pm \infty }\frac{2\kappa |\xi |^{2\alpha -1}}{\rho }\,\int _{0}^{|\xi |\delta }\frac{\cos \tau }{\tau ^{2\alpha -1}}d\tau -|\xi |^{2\alpha -1}\left(\frac{2\kappa \,|\xi |^{\alpha }}{\rho \, \omega (\xi )}\,\int _{0}^{|\xi |\delta }\frac{\sin \tau }{\tau ^{2\alpha }}d\tau \right)^2\\&=&\liminf _{R\rightarrow +\infty }\frac{2\kappa \delta ^{1-2\alpha }}{\rho R^{1-2\alpha }}\,\int _{0}^{R}\tau ^{1-2\alpha }\cos \tau d\tau -0\\&=&\liminf _{R\rightarrow +\infty }\frac{2\kappa \delta ^{1-2\alpha }}{\rho R^{1-2\alpha }}\,\int _{0}^{R} \Big ( (\tau ^{1-2\alpha }\sin \tau )^{\prime }-(1-2\alpha )\tau ^{-2\alpha }\sin \tau \Big )d\tau \\&=&\liminf _{R\rightarrow +\infty }\frac{2\kappa \delta ^{1-2\alpha }}{\rho R^{1-2\alpha }}\left(R^{1-2\alpha }\sin R-(1-2\alpha )\int _{0}^{+\infty }\frac{\sin \tau }{\tau ^{2\alpha }}d\tau \right)\\&=&-\frac{2\kappa \delta ^{1-2\alpha }}{\rho }.\) 
Similarly, when \(\alpha \in \left(0,\frac{1}{2}\right)\) ,
\(\sqrt{ \frac{4\kappa }{\rho }\,\int _{0}^{+\infty }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau }\limsup _{\xi \rightarrow \pm \infty }|\xi |^{1+\alpha } \omega ^{\prime \prime }(\xi )&=&\frac{2\kappa \delta ^{1-2\alpha }}{\rho }.\) 
These observations give formula () in Theorem REF .

Decay estimates
In this section we prove the relevant spatial decay properties of the solution of
(REF ) which also will be useful in the next section to prove the existence of conserved quantities for the peridynamic problem.
Theorem 3.1 
For every given \(t\ge 0\) , the function \(\mathbb {R}\ni x\mapsto u(t,x)\)  in (REF )
belongs to the Schwartz Space.

Let
\(\varpi (\xi ):=\omega ^2(\xi )\) 
and \(\Psi \)  be an even real analytic function such thatStatements like (REF ) mean that for
all \(\ell \in \mathbb {N}\) ,
\( \sup _{r\in [1,+\infty )}|\Psi ^{(\ell )}(r)|<+\infty ,\) 
where \(\Psi ^{(\ell )}\)  denotes the derivative of \(\Psi \)  of order \(\ell \) .

\({\mbox{the derivatives of~$\Psi $ of any orderare bounded in~$[1,+\infty )$.}}\) 
Thus, by the analyticity of \(\Psi \) , for suitable coefficients \(\Psi _j\in \mathbb {R}\) , with \(|\Psi _j|\le \frac{C^j}{(2j)!}\)  for some \(C>0\) ,
we write
\( \Psi (r):=\sum _{j=0}^{+\infty }\Psi _j\, r^{2j}.\) 
Therefore, setting
\( \psi (r):=\sum _{j=0}^{+\infty }\Psi _j\, r^{j},\) 
we find that
\( {\mbox{$\psi $is also a real analytic function.}}\) 
By (REF ), we know that
\(\begin{split}&\varpi (\xi )=\frac{4\kappa }{\rho \,\delta ^{2\alpha }}\,\int _{0}^{1}\frac{1-\cos (\xi \delta z)}{z^{1+2\alpha }}dz=\frac{4\kappa }{\rho \,\delta ^{2\alpha }}\,\int _{0}^{1}\sum _{k=1}^{+\infty } \frac{(-1)^{k+1} (\xi \delta )^{2k} z^{2k-1-2\alpha }}{(2k)!}\,dz\\&\qquad \qquad \qquad =\frac{2\kappa }{\rho }\,\sum _{k=1}^{+\infty } \frac{(-1)^{k+1} \xi ^{2k}\delta ^{2(k-\alpha )} }{(k-\alpha )(2k)!} .\end{split}\) 
In particular, we have that
\({\mbox{$\varpi $ is also a real analytic function.}}\) 
By composition,
it follows that
\({\mbox{$\phi (\xi ):=\psi (\varpi (\xi ))$ is also real analytic.}}\) 
Notice that, by construction
\( \phi (\xi )=\sum _{j=0}^{+\infty }\Psi _j\, (\varpi (\xi ))^{j}=\sum _{j=0}^{+\infty }\Psi _j\, (\omega (\xi ))^{2j}=\Psi (\omega (\xi )).\) 
We claim that
\({\mbox{the derivatives of~$\varpi $ of any order divided by~$(1+|\xi |^{2\alpha })$ are bounded.}}\) 
To this aim, in view of (REF ), it suffices to show that, for all \(\ell \in \mathbb {N}\) 
and \(|\xi |\ge 1\) ,
\(|\varpi ^{(\ell )}(\xi )|\le C_\ell \,(1+|\xi |^{2\alpha -\ell }).\) 
For this, we first show that for every \(\ell \)  there exist \(C_\ell \in \mathbb {R}\) 
and a function \(F_\ell \in C^\infty ((1/2,+\infty ))\)  whose
derivatives of any order are bounded in \([1,+\infty )\)  such that, for every \(\xi \in [1,+\infty )\) ,
\(\varpi ^{(\ell )} (\xi )= F_\ell (\xi )+ C_\ell \,\xi ^{2\alpha -\ell } \,\int _{0}^{\xi \delta }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau .\) 
In this notation, \(C_\ell \)  and \(F_\ell \)  may also depend on the structural parameters \(\kappa \) , \(\rho \)  and \(\delta \) ,
which are supposed to be fixed quantities.
Thus, we argue by induction over \(\ell \) . When \(\ell =0\) ,
changing variable \(\tau :=\xi \delta z\)  in the first integral of (REF ) we find that
\(\varpi (\xi )=\frac{4\kappa \xi ^{2\alpha }}{\rho }\,\int _{0}^{\xi \delta }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau ,\) 
which is (REF ) with \(C_0:=\frac{4\kappa }{\rho }\)  and \(F_0:=0\) .
We now suppose recursively that (REF ) holds true for the index \(\ell \) 
and we establish it for the index \(\ell +1\) . For this, taking one further derivative,
the inductive assumption leads to
\(\varpi ^{(\ell +1)} (\xi )&=& \frac{d}{d\xi }\left[F_\ell (\xi )+ C_\ell \,\xi ^{2\alpha -\ell } \,\int _{0}^{\xi \delta }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau \right]\\&=&F_\ell ^{\prime } (\xi )+ C_\ell \delta \xi ^{2\alpha -\ell } \frac{1-\cos (\xi \delta )}{(\xi \delta )^{1+2\alpha }}+ (2\alpha -\ell )C_\ell \,\xi ^{2\alpha -\ell -1} \,\int _{0}^{\xi \delta }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau \\&=&F_\ell ^{\prime } (\xi )+ C_\ell \delta ^{-2\alpha }\xi ^{-(\ell +1)} \big (1-\cos (\xi \delta )\big )+ (2\alpha -\ell )C_\ell \,\xi ^{2\alpha -(\ell +1)} \,\int _{0}^{\xi \delta }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau .\) 
Hence, we define \(C_{\ell +1}:=(2\alpha -\ell )C_\ell \)  and
\(F_{\ell +1}(\xi ):=F_\ell ^{\prime } (\xi )+ C_\ell \delta ^{-2\alpha }\xi ^{-(\ell +1)} \big (1-\cos (\xi \delta )\big ),\) 
and we stress that all the derivatives of \(F_{\ell +1}\)  are bounded in \([1,+\infty )\) ,
since so are the ones of \(F_\ell \) . The inductive step is thereby complete and we have thus
established (REF ).
We also notice that, if \(\xi \ge 1\) , then
\( \xi ^{2\alpha }\int _{0}^{\xi \delta }\frac{1-\cos \tau }{\tau ^{1+2\alpha }}d\tau \le \xi ^{2\alpha }\left[\int _{0}^{\delta }\frac{ \tau ^2}{\tau ^{1+2\alpha }}d\tau +\int _{\delta }^{\xi \delta }\frac{2}{\tau ^{1+2\alpha }}d\tau \right]\le \xi ^{2\alpha }\left[\frac{\delta ^{2-2\alpha }}{2-2\alpha }+\frac{ \delta ^{-2\alpha }}{2\alpha }\right]\le C \xi ^{2\alpha } ,\) 
for some \(C>0\)  depending on \(\delta \) . This and (REF )
yield that
\( |\varpi ^{(\ell )} (\xi )|\le |F_\ell (\xi )|+ C\,C_\ell \,\xi ^{2\alpha -\ell }.\) 
From this and the parity of \(\varpi \)  we obtain (REF )
(up to renaming \(C_\ell \) )
and thus (REF ).
Now we show that
\({\mbox{the derivatives of~$\psi $ of any order are bounded in~$[0,+\infty )$.}}\) 
From (REF ), it suffices to check this claim in \([1,+\infty )\) .
For this, using (REF ) and (REF ), we observe that \(\psi (r)=\Psi (\sqrt{r})\) 
for all \(r>0\) .
For this reason, by the Faà di Bruno's Formula,
\(\psi ^{(\ell )}(r)={d^{\ell } \over d\xi ^{\ell }}\big (\Psi (\sqrt{r})\big )=\sum {\frac{\ell !}{m!}}\; \Psi ^{(m_{1}+\cdots +m_{\ell })}(\sqrt{r})\;\prod _{j=1}^{\ell }\left(\frac{\frac{d^j}{dr^j}\sqrt{r}}{j!}\right)^{m_{j}},\) 
with the sum above ranging over all \(m\in \mathbb {N}^\ell \)  satisfying the constraint
\(\sum _{j=1}^\ell j\, m_j=\ell \) 
and the standard multiindex notation \(m!=m_{1}!\,m_{2}!\,\cdots \,m_{\ell }!\)  has been used.
We also observe the recursive fact that, for every \(j\ge 1\) ,
\( \frac{d^j}{dr^j}\sqrt{r}=\frac{(-1)^{j+1} r^{\frac{1-2j}{2}} }{2^j}\prod _{k=1}^{j-2} (2k+1).\) 
As a consequence, for all \(r\ge 1\)  and \(j\ge 1\) ,
\( \left|\frac{d^j}{dr^j}\sqrt{r}\right|\le \frac{r^{\frac{1-2j}{2}} }{2^j}\prod _{k=1}^{j-2} (2k+1)\le \frac{r^{\frac{1-2j}{2}} }{2^{2}}\prod _{k=1}^{j-2} (k+1)=\frac{r^{\frac{1-2j}{2}} \,(j-1)!}{4}\le \frac{(j-1)!}{4}\) 
and therefore
\(\prod _{j=1}^{\ell }\left(\frac{\frac{d^j}{dr^j}\sqrt{r}}{j!}\right)^{m_{j}}\le \prod _{j=1}^{\ell }\left(\frac{1}{4j}\right)^{m_{j}}\le \prod _{j=1}^{\ell }\left(\frac{1}{4}\right)^{m_{j}}\le C(\ell ),\) 
for some \(C(\ell )>0\) .
Hence,
exploiting (REF ) and (REF ),
and denoting by \(\bar{C}_\ell \)  a bound in \([1,+\infty )\)  for the derivatives of \(\Psi \) 
up to order \(\ell \)  (that takes into account the previous \(C(\ell )\)  too),
\( |\psi ^{(\ell )}(r)|\le \bar{C}_\ell \,\sum {\frac{\ell !}{m!}},\) 
from which the desired result in (REF ) plainly follows.
We now claim that for every \(\ell \in \mathbb {N}\)  there exists \(C^\#_\ell \ge 1\)  such that, for every \(\xi \in \mathbb {R}\) ,
\(|\phi ^{(\ell )}(\xi )|\le C^\#_\ell \,(1+|\xi |)^{C^\#_\ell }.\) 
To this end,
we exploit the Faà di Bruno's Formula to see that,
for every \(\ell \in \mathbb {N}\) ,
\(\phi ^{(\ell )}(\xi )={d^{\ell } \over d\xi ^{\ell }}\big (\psi (\varpi (\xi ))\big )=\sum {\frac{\ell !}{m!}}\; \psi ^{(m_{1}+\cdots +m_{\ell })}(\varpi (\xi ))\;\prod _{j=1}^{\ell }\left({\frac{\varpi ^{(j)}(\xi )}{j!}}\right)^{m_{j}},\) 
with the sum above ranging over all \(m\in \mathbb {N}^\ell \)  satisfying the constraint in (REF ).
Thus, recalling (REF )
and (REF ), we pick \(C^\star _\ell \ge 1\)  sufficiently large such that,
for all \(j\le \ell \) ,
\(&&|\varpi ^{(j)}(\xi )|\le {C^\star _\ell }\,(1+|\xi |^{2\alpha })\qquad {\mbox{ for every }}\xi \in \mathbb {R}\\{\mbox{and }}&&|\psi ^{(j)}(r)|\le {C^\star _\ell }\qquad {\mbox{ for every }}r\in [0,+\infty ).\) 
In this way, we find that
\(&& |\phi ^{(\ell )}(\xi )|\le C^\star _\ell \sum {\frac{\ell !}{m!}}\; \prod _{j=1}^{\ell }\left({\frac{{C^\star _\ell }\,(1+|\xi |^{2\alpha })}{j!}}\right)^{m_{j}}\\&&\qquad \le C^\star _\ell \sum {\frac{\ell !}{m!}}\; \prod _{j=1}^{\ell }\left({{{C^\star _\ell }\,(1+|\xi |^{2\alpha })}}\right)^{\ell }=(C^\star _\ell )^{1+\ell ^2}(1+|\xi |^{2\alpha })^{\ell ^2}\sum {\frac{\ell !}{m!}},\) 
which leads to (REF ).
Thus, from (REF ), (REF )
and (REF ), we obtain that
\(\begin{split}&{\mbox{if~$v$ isa smooth and rapidly decreasing function in the Schwartz Space,}}\\&{\mbox{then so is the function~$\xi \mapsto \hat{v}(\xi )\phi (\xi )=\hat{v}(\xi )\Psi (\omega (\xi ))$,}}\\&{\mbox{and so is its Fourier antitransform.}}\end{split}\) 
Applying this with \(\Psi (r):=\cos r\)  and with \(\Psi (r):=\frac{\sin r}{r}\) ,
and recalling the representation formula in (REF ),
we obtain the desired result in Theorem REF .
However, to perform this last step, we need to check that
the functions \(\Psi (r):=\cos r\)  and \(\Psi (r):=\frac{\sin r}{r}\) ,
satisfy the hypothesis in (REF ). This is obvious if \(\Psi (r):=\cos r\) .
If instead \(\Psi (r):=\frac{\sin r}{r}\) , we use that
\(\Psi (r)=\sum _{j=1}^{+\infty } \frac{(-1)^j r^{2j}}{(2j)!}\) 
to see that \(\Psi \)  is analytic, hence all its derivatives are bounded in \((-1,1)\) .
Thus, it only remains to check that all its derivatives are bounded in \(\mathbb {R}\setminus (-1,1)\) .
By even parity, it suffices to focus on \([1,+\infty )\) .
For this, we observe that \(\Psi =\Psi _1\,\Psi _2\) , where \(\Psi _1(r):=\sin r\) 
and \(\Psi _2(r):=1/r\) . Notice that the derivatives of \(\Psi _1\)  and \(\Psi _2\) 
of any order are bounded in \([1,+\infty )\) . This fact and the General Leibniz Rule
yield that
the derivatives of \(\Psi \)  of all orders are bounded in \([1,+\infty )\) ,
and the proof of Theorem REF  is thereby complete.
As a byproduct of the previous results, we now point out some integrability estimates
that will be used in Section  to introduce some useful conserved quantities
for solutions of equation (REF ).
Corollary 3.2 Let \(v_0,\,v_1\in \mathcal {S}(\mathbb {R})\)  and \(0<\alpha <1\) . Let \(u\)  be a solution of
problem (REF ). Let also \(w(t,x):=(1+|x|)u(t,x)\)  and \(W(t,x):=(1+|x|)u_t(t,x)\) .
Then, for all \(t>0\) ,
\(&& u_t(t,\cdot )\in L^2(\mathbb {R}),\\&& w(t,\cdot )\in L^1(\mathbb {R}),\\{\mbox{and }} && W(t,\cdot )\in L^1(\mathbb {R}).\) 
Moreover,
\(&&\int _{\mathbb {R}}\int _{-\delta }^{\delta }\frac{|u(t,x)-u(t,x-y)|^2}{|y|^{1+2\alpha }}\,dx\,dy<+\infty ,\\&&\nonumber \lim _{t\rightarrow 0^+}\int _{\mathbb {R}}\int _{-\delta }^{\delta }\frac{|u(t,x)-u(t,x-y)|^2}{|y|^{1+2\alpha }}\,dx\,dy=\frac{\rho }{\kappa }\int _{\mathbb {R}}\omega ^2(\xi )|\widehat{v_0}(\xi )|^2\,d\xi \\&&\qquad \qquad =\int _{\mathbb {R}}\int _{-\delta }^{\delta }\frac{|v_0(x)-v_0(x-y)|^2}{|y|^{1+2\alpha }}\,dx\,dy,\\&&\int _{\mathbb {R}}\int _{-\delta }^{\delta }\frac{|x|\,|2u(t,x)-u(t,x+y)-u(t,x-y)|}{|y|^{1+2\alpha }}\,dx\,dy<+\infty \) 
andHere and in the rest of this paper, the notation \(\int _A\int _B f(x,y)\,dx\,dy\) 
means that we are integrating over \(x\in A\)  and \(y\in B\)  (not vice-versa).
\(&&\lim _{t\rightarrow 0^+}\int _\mathbb {R}x u_t(t,x) \,dx = \int _\mathbb {R}x v_1(x)\, dx.\) 

From equation (REF ), we have that
\(u_t(t,x)=\int _{\mathbb {R}}\,e^{-i\xi x}\left[-\omega (\xi )\widehat{v_0}(\xi )\sin \left(\omega (\xi )\,t\right)+\widehat{v_1}(\xi )\cos \left(\omega (\xi )\,t\right)\right]d\xi \,.\) 
Then, using \(*\)  to denote complex conjugation and \(\delta (\cdot )\)  to denote the Dirac Delta
FunctionIt is useful
to recall that
\( \int _\mathbb {R}\int _\mathbb {R}e^{i\xi x}\, \psi (x)\,\widehat{\phi }(\xi )\,dx\,d\xi =2\pi \int _\mathbb {R}\widehat{\psi }(\xi )\,\widehat{\phi }(\xi )\,d\xi .\) 
Also, the Fourier transform of the convolution between \(\psi \)  and \(\phi \)  is
\(&&\widehat{\psi *\phi }(\xi )=\frac{1}{2\pi }\int _\mathbb {R}(\psi *\phi )(x) e^{ix\xi }\,dx=\frac{1}{2\pi }\int _\mathbb {R}\int _\mathbb {R}\psi (x-y)\phi (y) e^{ix\xi }\,dx\,dy\\&&\qquad \qquad \qquad =\frac{1}{2\pi }\int _\mathbb {R}\int _\mathbb {R}\psi (t)\phi (y) e^{it\xi } e^{iy\xi }\,dt\,dy=2\pi \widehat{\psi }(\xi )\,\widehat{\phi }(\xi ).\) 
From these observations and the inversion formula (recall footnote REF ), we find that
\( \int _\mathbb {R}\int _\mathbb {R}e^{i\xi x}\, \psi (x)\,\widehat{\phi }(\xi )\,dx\,d\xi =\int _\mathbb {R}\widehat{\psi *\phi }(\xi )\,d\xi =\int _\mathbb {R}e^{-i\xi 0}\,\widehat{\psi *\phi }(\xi )\,d\xi =\psi *\phi (0)=\int _\mathbb {R}\psi (x)\,\phi (x)\,dx.\) 
Hence, taking \(\psi :=1\)  and arguing in the sense of distributions,
\( \int _\mathbb {R}\int _\mathbb {R}e^{i\xi x}\, \widehat{\phi }(\xi )\,dx\,d\xi =\int _\mathbb {R}\phi (x)\,dx=\int _\mathbb {R}e^{i 0 x}\,\phi (x)\,dx=2\pi \widehat{\phi }(0)=2\pi \int _\mathbb {R}\delta (\xi )\,\widehat{\phi }(\xi )\,d\xi ,\) 
that is, distributionally,
\(\int _\mathbb {R}e^{i\xi x} \,dx=2\pi \delta (\xi ).\) 
 it follows that
\(\Vert u_t(t,\cdot ) \Vert ^2_{L^2(\mathbb {R})}=&\int _{\mathbb {R}}\int _{\mathbb {R}}\int _{\mathbb {R}} e^{-i(\xi -q)x}\left[-\omega (\xi )\widehat{v_0}(\xi )\sin \left(\omega (\xi )\,t\right)+\widehat{v_1}(\xi )\cos \left(\omega (\xi )\,t\right)\right]\nonumber \\&\times \left[-\omega (q)\widehat{v}^*_0(q)\sin \left(\omega (q)\,t\right)+\widehat{v}^*_1(q)\cos \left(\omega (q)\,t\right)\right]dx\,d\xi \,dq\nonumber \\=&2\pi \int _{\mathbb {R}}\int _{\mathbb {R}} \delta (\xi -q)\left[-\omega (\xi )\widehat{v_0}(\xi )\sin \left(\omega (\xi )\,t\right)+\widehat{v_1}(\xi )\cos \left(\omega (\xi )\,t\right)\right]\nonumber \\&\times \left[-\omega (q)\widehat{v}^*_0(q)\sin \left(\omega (q)\,t\right)+\widehat{v}^*_1(q)\cos \left(\omega (q)\,t\right)\right]d\xi \,dq\nonumber \\=&2\pi \int _{\mathbb {R}}\left[-\omega (\xi )\widehat{v_0}(\xi )\sin \left(\omega (\xi )\,t\right)+\widehat{v_1}(\xi )\cos \left(\omega (\xi )\,t\right)\right]\nonumber \\&\times \left[-\omega (\xi )\widehat{v}^*_0(\xi )\sin \left(\omega (\xi )\,t\right)+\widehat{v}^*_1(\xi )\cos \left(\omega (\xi )\,t\right)\right]d\xi \nonumber \\=&2\pi \int _{\mathbb {R}} \left\lbrace \omega ^2(\xi )|\widehat{v_0}(\xi )|^2\sin ^2\left(\omega (\xi )\,t\right)+|\widehat{v_1}(\xi )|^2\cos ^2\left(\omega (\xi )\,t\right)\right.\nonumber \\&\left.-\omega (\xi )\sin \left(\omega (\xi )\,t\right)\cos \left(\omega (\xi )\,t\right)\left[ \widehat{v_0}(\xi )\widehat{v}^*_1(\xi )+\widehat{v}^*_0(\xi )\widehat{v_1}(\xi ) \right]\right\rbrace d\xi .\) 

From this and the bound on \(\omega \)  in (REF ), we obtain the desired result in (REF ).
Also, the claim in () follows directly from Theorem REF .
Additionally, we have that
\( r\sin r=\sum _{j=0}^{+\infty }\frac{(-1)^j r^{2j+2}}{(2j+1)!}\) 
and therefore, given \(t>0\) ,
recalling the notation in (REF )
and the result in (REF ),
\( Z(t,\xi ):=\omega (\xi )\sin (\omega (\xi )t)=\frac{\omega (\xi )t\sin (\omega (\xi )t)}{t}=\sum _{j=0}^{+\infty }\frac{(-1)^j t^j (\varpi (\xi ))^{j+1}}{(2j+1)!},\) 
which is a real analytic function in the variable \(\xi \) .
Also, in view of (REF ),
we know that \(Z\)  grows at most polynomially at infinity in \(\xi \) ,
whence, if \(v\)  belongs to the Schwartz Space, then also
the function
\({\mbox{$\xi \mapsto \hat{v}(\xi ) Z(t,\xi )$ belongs to the Schwartz Space.}}\) 
Moreover, using (REF ),
we have that if \(v\)  belongs to the Schwartz Space, then also
the function
\({\mbox{$\xi \mapsto \hat{v}(\xi )\cos (\omega (\xi )t)$belongs to the Schwartz Space.}}\) 
Combining (REF ), (REF ) and (REF )
we obtain (), as desired.
Furthermore,
from equation (REF ), we have that
\(u(t,x)-u(t,x-y)=&\int _{\mathbb {R}} e^{-i\xi x}(1-e^{i\xi y})\left[\widehat{v_0}(\xi )\cos \left(\omega (\xi )\,t\right)+\frac{\widehat{v_1}(\xi )}{\omega (\xi )}\sin \left(\omega (\xi )\,t\right)\right]d\xi \nonumber \\=&\int _{\mathbb {R}} e^{-i\xi x}(1-\cos ( \xi y)-i\sin ( \xi y))\left[\widehat{v_0}(\xi )\cos \left(\omega (\xi )\,t\right)+\frac{\widehat{v_1}(\xi )}{\omega (\xi )}\sin \left(\omega (\xi )\,t\right)\right]d\xi \,.\) 

Hence, we obtain
\(\int _{\mathbb {R}}\int _{-\delta }^{\delta }&\frac{|u(t,x)-u(t,x-y)|^2}{|y|^{1+2\alpha }}dx\,dy=\int _{\mathbb {R}}\int _{-\delta }^{\delta }\frac{1}{|y|^{1+2\alpha }}\int _{\mathbb {R}} \int _{\mathbb {R}} e^{-i(\xi -q)x}\nonumber \\&\times (1-\cos (\xi y)-i\sin ( \xi y))(1-\cos ( qy)+i\sin ( qy))\nonumber \\&\times \left[\widehat{v_0}(\xi )\cos \left(\omega (\xi )\,t\right)+\frac{\widehat{v_1}(\xi )}{\omega (\xi )}\sin \left(\omega (\xi )\,t\right)\right]\left[\widehat{v}^*_0(q)\cos \left(\omega (q)\,t\right)+\frac{\widehat{v}^*_1(q)}{\omega (q)}\sin \left(\omega (q)\,t\right)\right]dx\,dy\,d\xi \,dq\nonumber \\=&2\pi \int _{-\delta }^{\delta }\frac{1}{|y|^{1+2\alpha }}\int _{\mathbb {R}} \int _{\mathbb {R}} \delta (\xi -q)(1-\cos ( \xi y)-i\sin ( \xi y))(1-\cos (qy)+i\sin ( qy))\nonumber \\&\times \left[\widehat{v_0}(\xi )\cos \left(\omega (\xi )\,t\right)+\frac{\widehat{v_1}(\xi )}{\omega (\xi )}\sin \left(\omega (\xi )\,t\right)\right]\left[\widehat{v}^*_0(q)\cos \left(\omega (q)\,t\right)+\frac{\widehat{v}^*_1(q)}{\omega (q)}\sin \left(\omega (q)\,t\right)\right]dy\,d\xi \,dq\nonumber \\=&2\pi \int _{-\delta }^{\delta }\frac{1}{|y|^{1+2\alpha }}\int _{\mathbb {R}}\left[(1-\cos (\xi y))^2+\sin ^2 \xi y\right]\nonumber \\&\times \left[\widehat{v_0}(\xi )\cos \left(\omega (\xi )\,t\right)+\frac{\widehat{v_1}(\xi )}{\omega (\xi )}\sin \left(\omega (\xi )\,t\right)\right]\left[\widehat{v}^*_0(\xi )\cos \left(\omega (\xi )\,t\right)+\frac{\widehat{v}^*_1(\xi )}{\omega (\xi )}\sin \left(\omega (\xi )\,t\right)\right]dy\,d\xi \nonumber \\=&4\pi \int _{\mathbb {R}} \int _{-\delta }^{\delta }\frac{1-\cos ( \xi y)}{|y|^{1+2\alpha }}\left\lbrace |\widehat{v_0}(\xi )|^2\cos ^2\left(\omega (\xi )\,t\right)+\frac{|\widehat{v_1}(\xi )|^2}{\omega ^2(\xi )}\sin ^2\left(\omega (\xi )\,t\right)\right.\nonumber \\&\left.\qquad \qquad \qquad +\frac{\cos \left( \omega (\xi )t \right)\sin \left( \omega (\xi )t \right)}{\omega (\xi )}\left[\widehat{v_0}(\xi )\widehat{v}^*_1(\xi )+\widehat{v_1}(\xi )\widehat{v}^*_0(\xi )\right]\right\rbrace d\xi \,dy\nonumber \\=&4\pi \,\delta ^{-2\,\alpha }\int _{\mathbb {R}}\int _{-1}^{1}\frac{1-\cos (\xi \delta z)}{|z|^{1+2\alpha }}\left\lbrace |\widehat{v_0}(\xi )|^2\cos ^2\left(\omega (\xi )\,t\right)+\frac{|\widehat{v_1}(\xi )|^2}{\omega ^2(\xi )}\sin ^2\left(\omega (\xi )\,t\right)\right.\nonumber \\&\left.\qquad \qquad \qquad +\frac{\cos \left( \omega (\xi )t \right)\sin \left( \omega (\xi )t \right)}{\omega (\xi )}\left[\widehat{v_0}(\xi )\widehat{v}^*_1(\xi )+\widehat{v_1}(\xi )\widehat{v}^*_0(\xi )\right]\right\rbrace d\xi \,dz\nonumber \\=&\frac{2\pi \rho }{\kappa }\int _{\mathbb {R}} \omega ^2(\xi )\left\lbrace |\widehat{v_0}(\xi )|^2\cos ^2\left(\omega (\xi )\,t\right)+\frac{|\widehat{v_1}(\xi )|^2}{\omega ^2(\xi )}\sin ^2\left(\omega (\xi )\,t\right)\right.\nonumber \\&\left. \qquad \qquad \qquad +\frac{\cos \left( \omega (\xi )t \right)\sin \left( \omega (\xi )t \right)}{\omega (\xi )}\left[\widehat{v_0}(\xi )\widehat{v}^*_1(\xi )+\widehat{v_1}(\xi )\widehat{v}^*_0(\xi )\right]\right\rbrace d\xi \nonumber \\=&\frac{2\pi \rho }{\kappa }\int _{\mathbb {R}}\left\lbrace \omega ^2(\xi )|\widehat{v_0}(\xi )|^2\cos ^2\left(\omega (\xi )\,t\right)+|\widehat{v_1}(\xi )|^2\sin ^2\left(\omega (\xi )\,t\right)\right.\nonumber \\&\left. \qquad \qquad \qquad +\omega (\xi )\cos \left( \omega (\xi )t \right)\sin \left( \omega (\xi )t \right)\left[\widehat{v_0}(\xi )\widehat{v}^*_1(\xi )+\widehat{v_1}(\xi )\widehat{v}^*_0(\xi )\right]\right\rbrace d\xi \,,\) 

where we have performed the change of variable \(z:=\delta y\)  and used (REF ). Thus,
recalling the bound on \(\omega \)  in (REF ) we obtain (REF ), as desired.
By taking the limit as \(t\rightarrow 0^+\)  in (REF ), we obtain the first
identity in (). Also, the second
identity in () follows by (REF ), the translation
invariance of the norm and
Plancherel Theorem; more precisely
\(&&\int _{\mathbb {R}}\int _{-\delta }^{\delta }\frac{|v_0(x)-v_0(x-y)|^2}{|y|^{1+2\alpha }}\,dx\,dy=\int _{\mathbb {R}}\int _{-\delta }^{\delta }\frac{v_0^2(x)+v_0^2(x-y)-2v_0(x)v_0(x-y)}{|y|^{1+2\alpha }}\,dx\,dy\\&&\qquad =2\int _{\mathbb {R}}\int _{-\delta }^{\delta }\frac{v_0^2(x)-v_0(x)v_0(x-y)}{|y|^{1+2\alpha }}\,dx\,dy=2\int _{\mathbb {R}}\int _{-\delta }^{\delta }\frac{|\widehat{v_0}(\xi )|^2-\widehat{v_0}(\xi )\,\overline{\widehat{v_0}(\xi )}\,e^{iy\xi }}{|y|^{1+2\alpha }}\,d\xi \, dy\\&&\qquad =2\int _{\mathbb {R}}\int _{-\delta }^{\delta }\frac{|\widehat{v_0}(\xi )|^2(1-e^{iy\xi })}{|y|^{1+2\alpha }}\,d\xi \,dy=\int _{\mathbb {R}}\int _{-\delta }^{\delta }\frac{|\widehat{v_0}(\xi )|^2(2-e^{-iy\xi }-e^{iy\xi })}{|y|^{1+2\alpha }}\,d\xi \, dy\\&&\qquad =2\int _{\mathbb {R}}\int _{-\delta }^{\delta }\frac{|\widehat{v_0}(\xi )|^2\big (1-\cos (y\xi )\big )}{|y|^{1+2\alpha }}\,d\xi \, dy=\frac{\rho }{\kappa }\int _{\mathbb {R}}\omega ^2(\xi )|\widehat{v_0}(\xi )|^2\,d\xi .\) 
Besides, for all \(y\in (-\delta ,\delta )\) ,
\(\begin{split}&|2u(t,x)-u(t,x+y)-u(t,x-y)|=\left|\int _0^{y} u_x(t,x+\theta )\,d\theta -\int _0^{y} u_x(t,x-\theta )\,d\theta \right|\\&\qquad =\left|\int _0^{y}\int _{-\theta }^{\theta } u_{xx}(t,x+\eta )\,d\theta \,d\eta \right|\le \sup _{\zeta \in (-\delta ,\delta )}|u_{xx}(t,x+\zeta )|\,y^2.\end{split}\) 
Also, by Theorem REF ,
\(&&(1+|x|)^3 \sup _{\zeta \in (-\delta ,\delta )}|u_{xx}(t,x+\zeta )|\le \sup _{\zeta \in (-\delta ,\delta )}(1+|x+\zeta |+\delta )^3 |u_{xx}(t,x+\zeta )|\\&&\qquad \le (1+\delta )^3\sup _{\zeta \in (-\delta ,\delta )}(1+|x+\zeta |)^3 |u_{xx}(t,x+\zeta )|\le C(1+\delta )^3,\) 
for some \(C>0\)  independent of \(x\) .
This and (REF ) lead to
\(&&\int _{\mathbb {R}}\int _{-\delta }^{\delta }\frac{|x|\,|2u(t,x)-u(t,x+y)-u(t,x-y)|}{|y|^{1+2\alpha }}\,dx\,dy\le C(1+\delta )^3\int _{\mathbb {R}}\int _{-\delta }^{\delta }\frac{|x|\,|y|^{1-2\alpha }}{(1+|x|)^3}\,dx\,dy\\&&\qquad \qquad =\frac{C(1+\delta )^3\delta ^{2-2\alpha }}{1-\alpha }\int _{\mathbb {R}} \frac{|x|}{(1+|x|)^3}dx,\) 
which is finite, thus proving ().
Now we use again (REF ), combined with the bound on \(\omega \)  obtained in (REF ),
to see that
\(\begin{split} &\left|\int _\mathbb {R}x u_t(t,x)\,dx-\int _\mathbb {R}x v_1(x)\,dx\right|\\=&\left|\int _\mathbb {R}\int _{\mathbb {R}}\,x e^{-i\xi x}\left[-\omega (\xi )\widehat{v_0}(\xi )\sin \left(\omega (\xi )\,t\right)+\widehat{v_1}(\xi )\cos \left(\omega (\xi )\,t\right)\right]\,d\xi \,dx-\int _\mathbb {R}x v_1(x)\,dx\right|\\\le &\left|\int _\mathbb {R}\int _{\mathbb {R}}\,\frac{d}{d\xi }(e^{-i\xi x})\, \omega (\xi )\widehat{v_0}(\xi )\sin \left(\omega (\xi )\,t\right)\,d\xi \,dx\right|\\&\qquad +\left|\int _\mathbb {R}\int _{\mathbb {R}}\,x e^{-i\xi x}\widehat{v_1}(\xi )\cos \left(\omega (\xi )\,t\right)\,d\xi \,dx-\int _\mathbb {R}\int _{\mathbb {R}}\,x e^{-i\xi x}\widehat{v_1}(\xi )\,d\xi \,dx\right|\\=&\left|\int _\mathbb {R}\int _{\mathbb {R}}\, e^{-i\xi x}\, \frac{d\Theta _1}{d\xi }(t,\xi )\,d\xi \,dx\right|+\left|\int _\mathbb {R}\int _{\mathbb {R}}\,\frac{d}{d\xi }( e^{-i\xi x})\,\widehat{v_1}(\xi )\Big (1-\cos \left(\omega (\xi )\,t\right)\Big )\,d\xi \,dx\right|\\=&\left|\int _\mathbb {R}\int _{\mathbb {R}}\, e^{-i\xi x}\, \frac{d\Theta _1}{d\xi }(t,\xi )\,d\xi \,dx\right|+\left|\int _\mathbb {R}\int _{\mathbb {R}}\, e^{-i\xi x}\,\frac{d\Theta _2}{d\xi }(t,\xi )\,d\xi \,dx\right|,\end{split}\) 
where
\( \Theta _1(t,\xi ):=\omega (\xi )\,\widehat{v_0}(\xi )\sin \left(\omega (\xi )\,t\right)\qquad {\mbox{and}}\qquad \Theta _2(t,\xi ):=\widehat{v_1}(\xi )\Big (1-\cos \left(\omega (\xi )\,t\right)\Big ).\) 
Recalling (REF ),
it is also convenient to consider the function
\([0,+\infty )\ni r\mapsto S(r):=\sqrt{r}\sin \sqrt{ r}=\sum _{j=0}^{+\infty }\frac{(-1)^j r^{j+1}}{(2j+1)!}\) 
We notice that \(S\)  can be extended to an analytic function defined
for all \(r\in \mathbb {R}\)  by using the series expansion in the right hand side of (REF ).
Also, for every \(\ell \in \mathbb {N}\)  and \(r>0\) ,
\( |S^{(\ell )}(r)|\le C_\ell \,(1+\sqrt{r}),\) 
for suitable \(C_\ell >0\) .
Indeed, if \(r\in [0,1]\)  this claim follows by taking derivatives
in the series expansion in (REF ), and if \(r\ge 1\) 
it follows from (REF ) and the General Leibniz Rule.
As a result, using (REF ), (REF )
and the Faà di Bruno's Formula, for every \(m\in \mathbb {N}\) ,
\(\left|\frac{d^m}{d\xi ^m}\big [\omega (\xi )t\,\sin \left(\omega (\xi )\,t\right)\big ]\right|=\left|\frac{d^m}{d\xi ^m} S\big (\varpi (\xi )t^2\big )\right|\le C^\star _m \,t^2\,(1+|\xi |)^{C^\star _m},\) 
for a suitable \(C^\star _m>0\) , and consequently,
using again the General Leibniz Rule and the fact that \(\widehat{v_0}\)  belongs to the Schwartz Space,
\(\left|\frac{d^m}{d\xi ^m}\Theta _1(t,\xi )\right|\le \frac{C^\#_m \,t}{1+\xi ^2},\) 
for some \(C^\#_m>0\) .
Integrating twice by parts in the variable \(\xi \)  when \(|x|>1\) , we thus find that
\(\begin{split}&\left|\int _\mathbb {R}\int _{\mathbb {R}}\, e^{-i\xi x}\, \frac{d\Theta _1}{d\xi }(t,\xi )\,d\xi \,dx\right|\\&\quad \le \left|\int _{\mathbb {R}\setminus [-1,1]} \int _{\mathbb {R}}\, e^{-i\xi x}\, \frac{d\Theta _1}{d\xi }(t,\xi )\,dx\,d\xi \right|+\left|\int _{[-1,1]} \int _{\mathbb {R}}\, e^{-i\xi x}\, \frac{d\Theta _1}{d\xi }(t,\xi )\,dx\,d\xi \right|\\&\quad \le \int _{\mathbb {R}\setminus [-1,1]} \int _{\mathbb {R}}\, \frac{C^\#_1 t}{1+\xi ^2}\,dx\,d\xi +\left|\int _{[-1,1]} \int _{\mathbb {R}}\,\frac{e^{-i\xi x}}{x^2}\,\frac{d^3\Theta _1}{d\xi ^3}(t,\xi )\,dx\,d\xi \right|\\&\quad \le O(t) +\int _{[-1,1]} \int _{\mathbb {R}}\,\frac{C^\#_3 t}{x^2\,(1+\xi ^2)}\,dx\,d\xi =O(t).\end{split}\) 
Similarly,
\( \left|\int _\mathbb {R}\int _{\mathbb {R}}\, e^{-i\xi x}\, \frac{d\Theta _2}{d\xi }(t,\xi )\,d\xi \,dx\right|=O(t).\) 
Plugging this and (REF ) into (REF )
we obtain (REF ) as desired.
We conclude this section with the following decay properties of the solutions of (REF ).
Theorem 3.3 
Let \(u(t,x)\)  be a solution of (REF ) according to (REF ).
Then
for all \(t>0\)  the following inequality holds true.
\(\Vert u(t,\cdot )\Vert _{L^2(\mathbb {R})}\le \left\Vert v_0\right\Vert _{L^2(\mathbb {R})}+\sqrt{2\pi }\left\Vert  \widehat{v_1}\min \left\lbrace  t,\,\frac{1}{\omega }\right\rbrace \right\Vert _{L^2(\mathbb {R})}\,.\) 

By (REF ) and Plancherel Theorem,
\(\begin{split}\frac{\Vert u(t,\cdot )\Vert ^2_{L^2(\mathbb {R})}}{2\pi }=&\int _\mathbb {R}\left| \widehat{v_0}(\xi )\cos \left(\omega (\xi )\,t\right)+\frac{\widehat{v_1}(\xi )}{\omega (\xi )}\sin \left(\omega (\xi )\,t\right)\right|^2\,d\xi \\\le &\int _\mathbb {R}\left( |\widehat{v_0}(\xi )|+\frac{|\widehat{v_1}(\xi )|}{\omega (\xi )}|\sin \left(\omega (\xi )\,t\right)|\right)^2\,d\xi \,.\end{split}\) 
Thus, since, for all \(r\in \mathbb {R}\) ,
\( \left|\frac{\sin r}{r}\right|\le \min \left\lbrace 1,\frac{1}{|r|}\right\rbrace ,\) 
we deduce from (REF ) that
\(\frac{\Vert u(t,\cdot )\Vert ^2_{L^2(\mathbb {R})}}{2\pi }\,\le \,\int _\mathbb {R}\left( |\widehat{v_0}(\xi )|+t\,|\widehat{v_1}(\xi )|\min \left\lbrace  1,\,\frac{1}{\omega (\xi )t}\right\rbrace \right)^2\,d\xi =\left\Vert |\widehat{v_0}|+|\widehat{v_1}|\min \left\lbrace  t,\,\frac{1}{\omega }\right\rbrace \right\Vert _{L^2(\mathbb {R})}^2\) 
and therefore
\(\frac{\Vert u(t,\cdot )\Vert _{L^2(\mathbb {R})}}{\sqrt{2\pi }}\,\le \,\left\Vert |\widehat{v_0}|+|\widehat{v_1}|\min \left\lbrace  t,\,\frac{1}{\omega }\right\rbrace \right\Vert _{L^2(\mathbb {R})}\le \,\left\Vert \widehat{v_0}\right\Vert _{L^2(\mathbb {R})}+\left\Vert  \widehat{v_1}\min \left\lbrace  t,\,\frac{1}{\omega }\right\rbrace \right\Vert _{L^2(\mathbb {R})},\) 
from which we obtain (REF )
using again Plancherel Theorem.
Theorem 3.4 
Let \(u(t,x)\)  be a solution of the (REF ) according
to (REF ).
Then, for every
\((t,x)\in \mathbb {R}_+\times \mathbb {R}\) ,
\(\begin{split}|u(t,x)|\,\le \,& \min \Bigg \lbrace \left\Vert \widehat{v_0}\right\Vert _{L^1(\mathbb {R})}+\left\Vert \widehat{v_1}\min \left\lbrace t,\frac{1}{\omega }\right\rbrace \right\Vert _{L^1(\mathbb {R})},\\&\frac{1+|x|}{t}\left(\left\Vert \frac{\widehat{v_0}}{\omega ^{\prime }}\right\Vert _{W^{1,1}((-\infty ,0))}+\left\Vert \frac{\widehat{v_0}}{\omega ^{\prime }}\right\Vert _{W^{1,1}((0,+\infty ))}+\left\Vert \frac{\widehat{v_1}}{\omega \omega ^{\prime }}\right\Vert _{W^{1,1}((-\infty ,0))}+\left\Vert \frac{\widehat{v_1}}{\omega \omega ^{\prime }}\right\Vert _{W^{1,1}((0,+\infty ))}\right)\Bigg \rbrace \,.\end{split}\) 

From
\(|u(t,x)|=\left|\int _{\mathbb {R}} e^{-i\xi x}\left[\widehat{v_0}(\xi )\cos \left(\omega (\xi )\,t\right)+\frac{\widehat{v_1}(\xi )}{\omega (\xi )}\sin \left(\omega (\xi )\,t\right)\right]d\xi \right|\) 
and (REF ) we deduce that
\(|u(t,x)|\le \int _{\mathbb {R}}|\widehat{v_0}(\xi )|+ |\widehat{v_1}(\xi )|\min \left\lbrace t,\frac{1}{\omega (\xi ) }\right\rbrace d\xi \,.\) 
Another consequence of (REF ) is that
\(|u(t,x)|=\left|\int _{\mathbb {R}} e^{-i\xi x}\left[\frac{\widehat{v_0}(\xi )}{\omega ^{\prime }(\xi )t}\partial _\xi \sin \left(\omega (\xi )\,t\right)-\frac{\widehat{v_1}(\xi )}{\omega (\xi )\,\omega ^{\prime }(\xi )t}\partial _\xi \cos \left(\omega (\xi )\,t\right)\right]d\xi \right|.\) 
Moreover, recalling (REF ) and (), we see that
\(\begin{split}&\int _{\mathbb {R}} e^{-i\xi x}\frac{\widehat{v_0}(\xi )}{\omega ^{\prime }(\xi )t}\partial _\xi \sin \left(\omega (\xi )\,t\right)d\xi \\=\,&\int _{-\infty }^0 e^{-i\xi x}\frac{\widehat{v_0}(\xi )}{\omega ^{\prime }(\xi )t}\partial _\xi \sin \left(\omega (\xi )\,t\right)d\xi +\int _{0}^{+\infty } e^{-i\xi x}\frac{\widehat{v_0}(\xi )}{\omega ^{\prime }(\xi )t}\partial _\xi \sin \left(\omega (\xi )\,t\right)d\xi \\=\,&\frac{ix}{t}\int _{-\infty }^0 e^{-i\xi x}\frac{\widehat{v_0}(\xi )}{\omega ^{\prime }(\xi )} \sin \left(\omega (\xi )\,t\right)d\xi +\frac{ix}{t}\int _{0}^{+\infty } e^{-i\xi x}\frac{\widehat{v_0}(\xi )}{\omega ^{\prime }(\xi )} \sin \left(\omega (\xi )\,t\right)d\xi \\&\qquad -\frac{1}{t}\int _{-\infty }^0 e^{-i\xi x}\left(\frac{\widehat{v_0}(\xi )}{\omega ^{\prime }(\xi ) }\right)^{\prime } \sin \left(\omega (\xi )\,t\right)d\xi -\frac{1}{t}\int _{0}^{+\infty } e^{-i\xi x}\left(\frac{\widehat{v_0}(\xi )}{\omega ^{\prime }(\xi ) } \right)^{\prime }\sin \left(\omega (\xi )\,t\right)d\xi \end{split}\) 
and therefore
\(\left|\int _{\mathbb {R}} e^{-i\xi x}\frac{\widehat{v_0}(\xi )}{\omega ^{\prime }(\xi )t}\partial _\xi \sin \left(\omega (\xi )\,t\right)\right|\le \frac{1+|x|}{t}\left(\left\Vert \frac{\widehat{v_0}}{\omega ^{\prime }}\right\Vert _{W^{1,1}((-\infty ,0))}+\left\Vert \frac{\widehat{v_0}}{\omega ^{\prime }}\right\Vert _{W^{1,1}((0,+\infty ))}\right).\) 
Additionally,
\(&&\int _{\mathbb {R}} e^{-i\xi x}\frac{\widehat{v_1}(\xi )}{\omega (\xi )\,\omega ^{\prime }(\xi )t}\partial _\xi \cos \left(\omega (\xi )\,t\right)d\xi \\&=&\int _{-\infty }^0 e^{-i\xi x}\frac{\widehat{v_1}(\xi )}{\omega (\xi )\,\omega ^{\prime }(\xi )t}\partial _\xi \cos \left(\omega (\xi )\,t\right)d\xi +\int _0^{+\infty } e^{-i\xi x}\frac{\widehat{v_1}(\xi )}{\omega (\xi )\,\omega ^{\prime }(\xi )t}\partial _\xi \cos \left(\omega (\xi )\,t\right)d\xi \\&=&\frac{ix}{t}\int _{-\infty }^0 e^{-i\xi x}\frac{\widehat{v_1}(\xi )}{\omega (\xi )\,\omega ^{\prime }(\xi )} \cos \left(\omega (\xi )\,t\right)d\xi +\frac{ix}{t}\int _0^{+\infty } e^{-i\xi x}\frac{\widehat{v_1}(\xi )}{\omega (\xi )\,\omega ^{\prime }(\xi )} \cos \left(\omega (\xi )\,t\right)d\xi \\&&\quad -\int _{-\infty }^0 e^{-i\xi x}\left(\frac{\widehat{v_1}(\xi )}{\omega (\xi )\,\omega ^{\prime }(\xi )}\right)^{\prime }\cos \left(\omega (\xi )\,t\right)d\xi +\int _0^{+\infty } e^{-i\xi x}\left(\frac{\widehat{v_1}(\xi )}{\omega (\xi )\,\omega ^{\prime }(\xi ) }\right)^{\prime }\cos \left(\omega (\xi )\,t\right)d\xi \) 
and, as a consequence,
\(\left|\int _{\mathbb {R}} e^{-i\xi x}\frac{\widehat{v_1}(\xi )}{\omega (\xi )\,\omega ^{\prime }(\xi )t}\partial _\xi \cos \left(\omega (\xi )\,t\right)d\xi \right|\le \frac{1+|x|}{t}\left(\left\Vert \frac{\widehat{v_1}}{\omega \omega ^{\prime }}\right\Vert _{W^{1,1}((-\infty ,0))}+\left\Vert \frac{\widehat{v_1}}{\omega \omega ^{\prime }}\right\Vert _{W^{1,1}((0,+\infty ))}\right).\) 
Owing to (REF ),
the latter estimate
and (REF ) entail that
\(|u(t,x)|\le \frac{1+|x|}{t}\left(\left\Vert \frac{\widehat{v_0}}{\omega ^{\prime }}\right\Vert _{W^{1,1}((-\infty ,0))}+\left\Vert \frac{\widehat{v_0}}{\omega ^{\prime }}\right\Vert _{W^{1,1}((0,+\infty ))}+\left\Vert \frac{\widehat{v_1}}{\omega \omega ^{\prime }}\right\Vert _{W^{1,1}((-\infty ,0))}+\left\Vert \frac{\widehat{v_1}}{\omega \omega ^{\prime }}\right\Vert _{W^{1,1}((0,+\infty ))}\right)\,.\) 
Thus, recalling (REF ),
we obtain the
desired result in (REF ).
Remark 3.5 We stress that
if \(v_0\)  and \(v_1\)  belong to the Schwartz Space, then in particular
\(\left\Vert \widehat{v_0}\right\Vert _{L^1(\mathbb {R})}+\left\Vert \widehat{v_1}\min \left\lbrace t,\frac{1}{\omega }\right\rbrace \right\Vert _{L^1(\mathbb {R})}<+\infty \) 
and consequently the right hand side of (REF )
is finite.
Furthermore, in light of (REF ), (REF ), (REF ),
(), (), () and (),
\( \frac{1}{\omega |\omega ^{\prime }|}+\frac{1}{\omega ^2|\omega ^{\prime }|}+\frac{|\omega ^{\prime \prime }|}{\omega (\omega ^{\prime })^2}=O\left(\frac{1}{|\xi | }+\frac{1}{\xi ^2 }+1\right)=O\left(\frac{1}{\xi ^2 }\right)\) 
as \(\xi \rightarrow 0^\pm \)  and
\(&& \frac{1}{\omega |\omega ^{\prime }|}+\frac{1}{\omega ^2|\omega ^{\prime }|}+\frac{|\omega ^{\prime \prime }|}{\omega (\omega ^{\prime })^2}=O\left(\frac{1}{|\xi |^{2\alpha -1}}+\frac{1}{|\xi |^{3\alpha -1}}+\frac{|\xi |^{\max \lbrace 2\alpha -1,0\rbrace -1-\alpha }}{|\xi |^{3\alpha -2}}\right)\\&&\qquad =O\left(\frac{1}{|\xi |^{2\alpha -1}}+\frac{1}{|\xi |^{\min \lbrace 2\alpha ,4\alpha -1\rbrace }}\right)=O\left(\frac{1}{|\xi |^{2\alpha -1}} \right)\) 
as \(\xi \rightarrow \pm \infty \) .
By (REF ), it follows that additional
assumptions (beside being in the Schwartz Space)
must be taken on \(\widehat{v_1}\) 
if one wishes that
\(\left\Vert \frac{\widehat{v_1}}{\omega \omega ^{\prime }}\right\Vert _{W^{1,1}((-\infty ,0))}+\left\Vert \frac{\widehat{v_1}}{\omega \omega ^{\prime }}\right\Vert _{W^{1,1}((0,+\infty ))}<+\infty .\) 


Conserved quantities
In this section, we investigate the conservation properties of equation (REF ).
For this, we introduce the following definition:
Definition 4.1 
Let \(u(t,x)\)  be a solution of (REF ). We define the following functionals:
\(&\text{Energy}&\quad &E[u(t,\cdot )]:=\frac{\rho }{2}\Vert  u_t(t,\cdot ) \Vert ^2_{L^2(\mathbb {R})}+\frac{\kappa }{2}\int _{\mathbb {R}}\int _{-\delta }^{\delta }\frac{|u(t,x)-u(t,x-y)|^2}{|y|^{1+2\alpha }}\,dx\,dy\,,\\&\text{Momentum}&\quad &P[u(t,\cdot )]:= \rho \int _{\mathbb {R}} u_t(t,x)dx,\\&\text{Angular momentum}&\quad &L[u(t,\cdot )] := \rho \int _{\mathbb {R}}x\,u_t(t,x)dx.\) 

We stress that the definition in (REF ) is well posed thanks to (REF )
and (REF ). Similarly, the definitions in () and ()
are well posed thanks to ().
All the quantities introduced in Definition REF 
are conserved by the equation, according to the following result:
Theorem 4.2 
If \(u\)  is a solution of problem (REF ), then
\(i)\)
\(u\)  preserves Energy according to (REF ) in Definition REF ;
\(ii)\)
\(u\)  preserves Momentum according to () in Definition  REF
\(iii)\)
\(u\)  preserves Angular Momentum according to ()
in Definition REF .


Theorem REF  is actually the byproduct of the forthcoming Theorems REF , REF 
and REF .
Theorem 4.3 (Angular Momentum conservation) 
Let \(u(t,x)\)  be a solution of (REF ). Then
\(L[u(t,\cdot )] =\rho \int _{\mathbb {R}}x\,v_1(x)\,dx.\) 

Let \(u(t,x)\)  be a solution of (REF ). It follows that
\(\rho \frac{d}{dt}\int _{\mathbb {R}}x\,u_t\,dx=&\int _{\mathbb {R}}\rho \,x\,u_{tt}\,dx=-2\,\kappa \int _{\mathbb {R}}\int _{-\delta }^{\delta }x\frac{u(t,x)-u(t,x-y)}{|y|^{1+2\,\alpha }}dx\,dy\nonumber \\=&-2\kappa \int _{\mathbb {R}}\int _{-\delta }^\delta \left[x\,u(t,x)-\left(x-y\right)\,u(t,x-y)-y\,u(t,x-y)\right]dx\,\frac{dy}{|y|^{1+2\,\alpha }}\nonumber \\=&-2\kappa \lim _{\varepsilon \rightarrow 0,\,\varepsilon ^{\prime }\rightarrow 0}\int _{\left[-\delta ,- \varepsilon \right]\cup \left[ \varepsilon ^{\prime },\delta \right]}\underbrace{\left[ \int _{\mathbb {R}}x\,u(t,x)dx - \int _{\mathbb {R}}\left(x-y\right)\,u(t,x-y)dx \right]}_{=0}\frac{dy}{|y|^{1+2\,\alpha }}\nonumber \\&+2\kappa \int _{\mathbb {R}}\int _{-\delta }^\delta \frac{y}{|y|^{1+2\alpha }}\,u(t,x-y)dx\,dy\nonumber \\=&2\kappa \lim _{\varepsilon \rightarrow 0}\Bigg \lbrace  \int _{\varepsilon }^\delta \frac{1}{y^{2\alpha }}\left[ \int _{\mathbb {R}}u(t,x-y)dx \right]dy-\int _{-\delta }^{- \varepsilon } \frac{1}{|y|^{2\alpha }}\left[ \int _{\mathbb {R}}u(t,x-y)dx \right]dy\Bigg \rbrace \nonumber \\=&2\kappa \lim _{\varepsilon \rightarrow 0}\Bigg \lbrace  \int _{\varepsilon }^\delta \frac{1}{y^{2\alpha }}{\left[ \int _{\mathbb {R}}u(t,x-y)dx \right]}dy-\int _{ \varepsilon }^{\delta } \frac{1}{y^{2\alpha }}{\left[ \int _{\mathbb {R}}u(t,x+y)dx \right]}dy\Bigg \rbrace \nonumber \\=&2\kappa \lim _{\varepsilon \rightarrow 0}\Bigg \lbrace  \int _\varepsilon ^\delta \frac{1}{y^{2\alpha }}\underbrace{\left[ \int _{\mathbb {R}}u(t,x-y)dx-\int _{\mathbb {R}}u(t,x+y)dx \right]}_{=0}dy\Bigg \rbrace ,\) 

where the two integrals in the third and last lines cancel due to the translational invariance –
and we stress that the integrals involved in the computations are finite, thanks
to (), () and () (recall also (REF )).
We have thus shown that the angular momentum is constant in time,
whence (REF ) follows from (REF )
and ().
Theorem 4.4 (Energy conservation) 
Let \(u(t,x)\)  be a solution of the (REF ) according to (REF ). Then
\(\begin{split}E[u(t,\cdot )]\,&=\,\rho \pi \int _{\mathbb {R}} \left\lbrace \omega ^2(\xi )|\widehat{v_0}(\xi )|^2+|\widehat{v_1}(\xi )|^2\right\rbrace d\xi \\&=\,\frac{\rho \pi }{2}\Vert v_1\Vert ^2_{L^2(\mathbb {R})}+\frac{\kappa }{2}\int _{\mathbb {R}}\int _{-\delta }^{\delta }\frac{|v_0(x)-v_0(x-y)|^2}{|y|^{1+2\alpha }}\,dx\,dy.\end{split}\) 

From (REF ) in
Definition REF  and equations (REF ) and (REF ), we get
\(E[u(t,\cdot )]=&\rho \pi \int _{\mathbb {R}} \left\lbrace \omega ^2(\xi )|\widehat{v_0}(\xi )|^2\sin ^2\left(\omega (\xi )\,t\right)+|\widehat{v_1}(\xi )|^2\cos ^2\left(\omega (\xi )\,t\right)\right.\nonumber \\&\left.-\omega (\xi )\sin \left(\omega (\xi )\,t\right)\cos \left(\omega (\xi )\,t\right)\left[ \widehat{v_0}(\xi )\widehat{v}^*_1(\xi )+\widehat{v}^*_0(\xi )\widehat{v_1}(\xi ) \right]\right.\nonumber \\&+\omega ^2(\xi )|\widehat{v_0}(\xi )|^2\cos ^2\left(\omega (\xi )\,t\right)+|\widehat{v_1}(\xi )|^2\sin ^2\left(\omega (\xi )\,t\right)\nonumber \\&\left.+\omega (\xi )\cos \left( \omega (\xi )t \right)\sin \left( \omega (\xi )t \right)\left[\widehat{v_0}(\xi )\widehat{v}^*_1(\xi )+\widehat{v_1}(\xi )\widehat{v}^*_0(\xi )\right]\right\rbrace d\xi \nonumber \\=&\rho \pi \int _{\mathbb {R}} \left\lbrace \omega ^2(\xi )|\widehat{v_0}(\xi )|^2+|\widehat{v_1}(\xi )|^2\right\rbrace d\xi \,.\) 

This establishes the first identity in (REF ).
In particular, the energy is constant in time
and recalling () we obtain the second identity
in (REF ).
Theorem 4.5 (Momentum conservation) 
Let \(u(t,x)\)  be a solution of the (REF ) according to (REF ). Then
\(P[u(t,\cdot )]=\rho \,\int _{\mathbb {R}}v_1(x)\,dx.\) 

From (REF ) and (REF ) it follows that
\(P[u(t,\cdot )]=&2\rho \pi \int _{\mathbb {R}}\delta (0)\left[-\omega (\xi )\widehat{v_0}(\xi )\sin \left(\omega (\xi )\,t\right)+\widehat{v_1}(\xi )\cos \left(\omega (\xi )\,t\right)\right]\,d\xi \nonumber \\=&2\rho \pi \left[-\omega (0)\widehat{v_0}(0)\sin \left(\omega (0)\,t\right)+\widehat{v_1}(0)\cos \left(\omega (0)\,t\right)\right]=2\rho \pi \,\widehat{v_1}(0)\,.\) 

Numerics
From now on, let us consider the numerical integration for the case \(\alpha =1/10\) ,
\(\rho =1\)  and \(\kappa =1/2\) , whereas \(\delta \)  will be fixed case by case. Moreover, we fix initial conditions such that
\(v_0(x)=\sqrt{2\pi }\,e^{-2\,x^2}\qquad \text{and}\qquad v_1(x)=4\,v\,x\sqrt{2\pi }\,e^{-2\,x^2}\,,\) 
i.e. the initial deformation is Gaussian, with square root
of the variance \(\sigma =1/2\)  and the initial velocity is given by the initial condition of a traveling wave \(v_1(x)=v\,v_0^{\prime }(x)\) . The value of \(v\) 
will be specified case by case later as well. Thus, in Fourier Space, we have
\(\widehat{v_0}(\xi )=\frac{1}{2}e^{-\frac{\xi ^2}{8}}\qquad \text{and}\qquad \widehat{v_1}(\xi )=i\,v\,\xi \,\widehat{v_0}(\xi ).\) 
Notice that \(\widehat{v_0}(\xi )\)  is a Gaussian with \(\widehat{\sigma }=2\) .
The numerical evolution of this Gaussian
according to (REF ) and (REF )
is depicted in Figure REF , where \(\delta =1\)  and \(v=0\) .
Let us emphasize some important differences
exhibited in Figure REF  with respect to the
classical case of the wave equation (in which the solution
is simply the sum of two traveling positive Gaussians, as shown inThis numerical solution holds for the wave equation \(u_{tt}=u_{xx}\) , with initial conditions given by equation (REF ) with \(v=0\) . This leads to the analogous of (REF ) where \(\omega (\xi )\)  is replaced by \(\xi \) . Figure REF ).
First of all, the pattern in Figure REF  is that of a sign-changing
solutions. Also, multiple critical points happen to
arise as time goes.
Overall, in this situation,
with respect to the classical wave equation,
the peridynamic case seems to produce
additional oscillations.
Certainly, it is desirable to carry on
further analytical and numerical investigations of these possible
phenomena.
<FIGURE><FIGURE>Moreover, in Figure REF  we report numerical solutions for the Cauchy problem given by initial conditions (REF ) with \(v=1\)  and \(\delta =1\) .
<FIGURE>We notice the presence of secondary oscillations left behind the wavefront, whose amplitude slowly decreases as time evolves. These two features are not present in the solution of the classical equation \(u_{tt}=u_{xx}\) . Indeed, in this case only a single Gaussian is expected to travel without neither deformation or damping of the amplitude.
Finally, in Figure REF  we show numerical solutions of the peridynamic model for three different values of \(\delta \)  and initial velocity given by \(v=\delta ^{1-\alpha }/\sqrt{2(1-\alpha )}\) . The latter choice is inspired by the limit of the dispersion relation on large scales, proven in Theorem REF .
<FIGURE>Figure REF  refers to \(\delta =5/2\)  (red), \(\delta =1\)  (green) and \(\delta =1/10\)  (blue). To properly interpret this numerical solution, we compare the values of \(\delta \)  with the dispersion of the Gaussian in the chosen initial condition (\(\sigma =1/2\) ). For \(\delta =5/2\) , we have that \(\delta = 5 \sigma \) . This means that the deformation introduced by the initial conditions involves scales which are small when compared with \(\delta \) , i.e. the characteristic range of the peridynamic model. This leads to a naive expectations that most of the modes \(\xi \)  propagates with dispersion relation of order \(\xi ^\alpha \)  and hence the evolution is highly dispersive. This expectation is in qualitative agreement with what shown in the red plots of Figure REF .
On the opposite case, when \(\delta =1/10\) , we have that \(\delta = \sigma /5\)  and then the deformation induced by the initial conditions are on large scales when compared with \(\delta \) . In this case, most of the involved scales propagates with dispersion relation \(\approx \delta ^{1-\alpha }/\sqrt{2(1-\alpha )}\,\xi \) , whose group velocity is the same as the one given in the initial conditions. Hence we expect that this case is nondispersive. Bottom panels of Figure REF  are in line with this expectation.
Finally, middle panels in Figure REF  exploit the case when \(\delta =1\)  and \(\sigma =1/2\)  are of the same order. We notice an evolution which is still dispersive, just as in Figure REF . However, in Figure REF  the secondary oscillations have smaller amplitude that in Figure REF . We address this behavior to the fact that, in Figure REF , deformation of large scales travel with velocity \(v=\delta ^{1-\alpha }/\sqrt{2(1-\alpha )}\)  which is just the one emerging from the dispersion relation when \(\xi \rightarrow 0\) .

appendix

Recovering the classical wave equation as \(\alpha \rightarrow 1^-\)
In this appendix we
discuss how
problem (REF )
recovers the classical wave equation as \(\alpha \rightarrow 1^-\) ,
and how the explicit solution provided
in (REF )
and (REF ) recovers in the limit
the one obtained for the wave
equation via Fourier methods.
Lemma 1.1 
If \(u\in C^2(\mathbb {R})\cap L^\infty (\mathbb {R})\) , then
\( \lim _{\alpha \rightarrow 1^-}(1-\alpha ) K(u)=C\kappa \Delta u,\) 
for a suitable constant \(C>0\) .

By the definition of \(K(u)\) 
in (REF ),
and exploiting [7]}, we have that for all \(\alpha \in (0,1)\) ,
\(-(-\Delta )^\alpha u=C_\alpha \int _{\mathbb {R}}\frac{u(t,x-y)-u(t,x)}{|y|^{1+2\,\alpha }}dy=C_\alpha \left[\frac{K(u)}{2\kappa }+\int _{\mathbb {R}\setminus (-\delta ,\delta )}\frac{u(t,x-y)-u(t,x)}{|y|^{1+2\,\alpha }}dy\right],\) 
for some \(C_\alpha \)  such that
\(\lim _{\alpha \rightarrow 1^-}\frac{C_\alpha }{1-\alpha }=C_\star ,\) 
for a suitable constant \(C_\star >0\) .
Furthermore,
\(\int _{\mathbb {R}\setminus (-\delta ,\delta )}\frac{|u(t,x-y)-u(t,x)|}{|y|^{1+2\,\alpha }}dy\le \int _{\mathbb {R}\setminus (-\delta ,\delta )}\frac{2\Vert u\Vert _{L^\infty (\mathbb {R})}}{|y|^{1+2\,\alpha }}dy= \int _{ \delta }^{+\infty }\frac{4\Vert u\Vert _{L^\infty (\mathbb {R})}}{y^{1+2\,\alpha }}dy=\frac{2\Vert u\Vert _{L^\infty (\mathbb {R})}}{\alpha \delta ^{2\alpha }}\) 
and consequently
\( \lim _{\alpha \rightarrow 1^-}(1-\alpha ) \int _{\mathbb {R}\setminus (-\delta ,\delta )}\frac{|u(t,x-y)-u(t,x)|}{|y|^{1+2\,\alpha }}dy=0.\) 
Using this, (REF ), (REF ) and [7]},
we conclude that
\(&&\Delta u=\lim _{\alpha \rightarrow 1^-}-(-\Delta )^\alpha u=\lim _{\alpha \rightarrow 1^-}\frac{C_\alpha }{1-\alpha }\,(1-\alpha )\left[\frac{K(u)}{2\kappa }+\int _{\mathbb {R}\setminus (-\delta ,\delta )}\frac{u(t,x-y)-u(t,x)}{|y|^{1+2\,\alpha }}dy\right]\\&&\qquad \qquad =\frac{C_\star }{2\kappa } \lim _{\alpha \rightarrow 1^-}(1-\alpha )K(u),\) 
as desired.
Lemma 1.2 
We have that
\( \lim _{\alpha \rightarrow 1^-}\sqrt{1-\alpha }\;\omega (\xi )=\frac{C\sqrt{\kappa } \,|\xi |}{\sqrt{\rho }},\) 
for a suitable constant \(C>0\) .

In view of the parity of \(\omega \) ,
we can suppose that \(\xi >0\) .
Also, by [7]}, we know that
\(\lim _{\alpha \rightarrow 1^-} (1-\alpha )\int _\mathbb {R}\frac{1-\cos \tau }{|\tau |^{1+2\alpha }}\,d\tau =C_\star ,\) 
for some \(C_\star >0\) .
Thus, from (REF )
\(\begin{split}\lim _{\alpha \rightarrow 1^-}(1-\alpha )\omega ^2(\xi )\,&=\lim _{\alpha \rightarrow 1^-}\frac{2(1-\alpha )\kappa }{\rho \,\delta ^{2\alpha }}\,\int _{-1}^{1}\frac{1-\cos (\xi \delta z)}{|z|^{1+2\alpha }}dz\\ &=\lim _{\alpha \rightarrow 1^-}\frac{2(1-\alpha )\kappa |\xi |^{2\alpha }}{\rho }\,\int _{-\xi \delta }^{\xi \delta }\frac{1-\cos (\tau )}{|\tau |^{1+2\alpha }}d\tau \\&=\frac{2\kappa C_\star \,|\xi |^{2}}{\rho }+\lim _{\alpha \rightarrow 1^-}\frac{4(1-\alpha )\kappa |\xi |^{2\alpha }}{\rho }\left(\int _{\xi \delta }^{+\infty }\frac{1-\cos (\tau )}{\tau ^{1+2\alpha }}d\tau \right).\end{split}\) 
Additionally,
\(\int _{\xi \delta }^{+\infty }\frac{1-\cos (\tau )}{\tau ^{1+2\alpha }}d\tau \le \int _{\xi \delta }^{+\infty }\frac{2}{\tau ^{1+2\alpha }}d\tau =\frac{(\xi \delta )^{2\alpha }}{\alpha }\) 
This and (REF ) entail that
\( \lim _{\alpha \rightarrow 1^-}(1-\alpha )\omega ^2(\xi )=\frac{2\kappa C_\star \,|\xi |^{2}}{\rho },\) 
from which the desired result follows.


An elementary proof of (REF )
We use an ad-hoc and rather delicate
modification of a classical
argument used in complex analysis (see e.g. [10]}).
For this we use a contour integration as in Figure REF ,
with \(\gamma =\gamma _1\cup \gamma _2\cup \gamma _3\cup \gamma _4\) ,
oriented counterclockwise.
<FIGURE>We observe that, by Cauchy's Theorem,
\(\int _\gamma \frac{e^{iz}-1}{z^{1+2\alpha }}\,dz=0\) 
and therefore
\(\lim _{{\epsilon \rightarrow 0^+}\atop {R\rightarrow +\infty }}\Re \left(\int _\gamma \frac{e^{iz}-1}{z^{1+2\alpha }}\,dz\right)=0.\) 
Thus, since
\( \lim _{{\epsilon \rightarrow 0^+}\atop {R\rightarrow +\infty }}\Re \left(\int _{\gamma _1} \frac{e^{iz}-1}{z^{1+2\alpha }}\,dz\right)=\lim _{{\epsilon \rightarrow 0^+}\atop {R\rightarrow +\infty }}\Re \left(\int _{\epsilon }^R \frac{e^{it}-1}{t^{1+2\alpha }}\,dt\right)=\lim _{{\epsilon \rightarrow 0^+}\atop {R\rightarrow +\infty }}\int _{\epsilon }^R \frac{\cos t-1}{t^{1+2\alpha }}\,dt=\int _{0}^{+\infty } \frac{\cos t-1}{t^{1+2\alpha }}\,dt,\) 
we deduce from (REF ) that
\(\int _{0}^{+\infty } \frac{1-\cos t }{t^{1+2\alpha }}\,dt=\lim _{{\epsilon \rightarrow 0^+}\atop {R\rightarrow +\infty }}\Re \left(\int _{\gamma _2\cup \gamma _3\cup \gamma _4} \frac{e^{iz}-1}{z^{1+2\alpha }}\,dz\right).\) 
We also point out that if \(z=x+iy\)  with \(x\in \mathbb {R}\)  and \(y\ge 0\)  then
\( |e^{iz}|=|e^{-y}|\le 1.\) 
Hence, since \(\gamma _2\) 
is the quarter of circle (traveled anticlockwise)
of the form \(\lbrace z=R e^{it},\;t\in (0,\pi /2)\rbrace \) ,
we have that if \(z\in \gamma _2\)  then \(\left|\frac{e^{iz}-1}{z^{1+2\alpha }}\right|=\frac{|e^{iz}-1|}{|z|^{1+2\alpha }}\le \frac{2}{R^{1+2\alpha }}\) .
Accordingly,
\(\left|\lim _{{\epsilon \rightarrow 0^+}\atop {R\rightarrow +\infty }}\Re \left( \int _{\gamma _2} \frac{e^{iz}-1}{z^{1+2\alpha }}\,dz\right)\right|\le \lim _{{R\rightarrow +\infty }}\frac{\pi R}{ R^{1+2\alpha }}=0.\) 
Now we note that \(\gamma _4\)  is the quarter of circle (traveled clockwise)
of the form \(\lbrace z=\epsilon e^{it},\;t\in (0,\pi /2)\rbrace \) .
Also, if \(z\in \gamma _4\) , for small \(\epsilon \)  we have that
\( \frac{e^{iz}-1}{z^{1+2\alpha }}=\frac{i}{z^{2\alpha }}+O(\epsilon ^{1-2\alpha })\) 
and therefore
\( \begin{split}&\Re \left( \int _{\gamma _4} \frac{e^{iz}-1}{z^{1+2\alpha }}\,dz\right)=\Re \left( \int _{\gamma _4} \frac{i}{z^{2\alpha }}\,dz\right)+O(\epsilon ^{2-2\alpha })=\epsilon ^{1-2\alpha }\Re \left( \int _0^{\pi /2} \frac{e^{it}}{e^{2i\alpha t}}\,dt\right)+O(\epsilon ^{2-2\alpha })\\&\qquad =\epsilon ^{1-2\alpha }\int _0^{\pi /2} \cos ((2\alpha -1) t)\,dt+O(\epsilon ^{2-2\alpha })=\frac{\epsilon ^{1-2\alpha }}{2\alpha -1}\sin \frac{(2\alpha -1)\pi }{2}+O(\epsilon ^{2-2\alpha })\\&\qquad =-\frac{\epsilon ^{1-2\alpha }}{2\alpha -1}\cos (\pi \alpha )+O(\epsilon ^{2-2\alpha }).\end{split}\) 
As for the integral on \(\gamma _3=\lbrace z=it,\;t\in (\epsilon ,R)\rbrace \)  (oriented
downwards), using that
\( i^{2\alpha }=(e^{\frac{i\pi }{2}})^{2\alpha }=e^{ {i\pi }\alpha }\) 
we have
\(\begin{split}&\Re \left(\int _{\gamma _3}\frac{e^{iz}-1}{z^{1+2\alpha }}\,dz\right)=-\Re \left(i\int _{\epsilon }^R\frac{e^{-t}-1}{i^{1+2\alpha } t^{1+2\alpha }}\,dt\right)=-\Re \left(e^{-i\pi \alpha }\int _{\epsilon }^R\frac{e^{-t}-1}{ t^{1+2\alpha }}\,dt\right)\\&\qquad =\frac{1}{2\alpha }\cos (\pi \alpha )\int _{\epsilon }^R\left[\frac{d}{dt}\left((e^{-t}-1)t^{-2\alpha }\right)+ e^{-t} t^{-2\alpha }\right]\,dt\\&\qquad =\frac{1}{2\alpha }\cos (\pi \alpha )\left[(e^{-R}-1)R^{-2\alpha }-(e^{-\epsilon }-1)\epsilon ^{-2\alpha }+\int _{\epsilon }^Re^{-t} t^{-2\alpha }\,dt\right]\\&\qquad =\frac{1}{2\alpha }\cos (\pi \alpha )\left[\epsilon ^{1-2\alpha }+\int _{\epsilon }^Re^{-t} t^{-2\alpha }\,dt\right]+O(R^{-2\alpha })+O(\epsilon ^{2-2\alpha })\\\\&\qquad =\frac{1}{2\alpha }\cos (\pi \alpha )\left[\epsilon ^{1-2\alpha }+\frac{1}{1-2\alpha }\int _{\epsilon }^R\left(\frac{d}{dt}\big (e^{-t} t^{1-2\alpha }\big )+e^{-t} t^{1-2\alpha }\right)\,dt\right]+O(R^{-2\alpha })+O(\epsilon ^{2-2\alpha })\\&\qquad =\frac{1}{2\alpha }\cos (\pi \alpha )\left[\epsilon ^{1-2\alpha }+\frac{1}{1-2\alpha }\left(e^{-R} R^{1-2\alpha }-e^{-\epsilon }\epsilon ^{1-2\alpha }+\int _{\epsilon }^Re^{-t} t^{1-2\alpha }\,dt\right)\right]+O(R^{-2\alpha })+O(\epsilon ^{2-2\alpha })\\&\qquad =\frac{1}{2\alpha }\cos (\pi \alpha )\left[\epsilon ^{1-2\alpha }+\frac{1}{1-2\alpha }\left(-\epsilon ^{1-2\alpha }+\int _{\epsilon }^Re^{-t} t^{1-2\alpha }\,dt\right)\right]+O(R^{-2\alpha })+O(\epsilon ^{2-2\alpha })\\&\qquad =\frac{1}{2\alpha (1-2\alpha )}\cos (\pi \alpha )\left[-2\alpha \epsilon ^{1-2\alpha }+\int _{\epsilon }^Re^{-t} t^{1-2\alpha }\,dt \right]+O(R^{-2\alpha })+O(\epsilon ^{2-2\alpha }).\end{split}\) 
As a result, recalling (REF ),
\(&&\lim _{{\epsilon \rightarrow 0^+}\atop {R\rightarrow +\infty }}\Re \left( \int _{\gamma _3\cup \gamma _4}\frac{e^{iz}-1}{z^{1+2\alpha }}\,dz\right)=\lim _{{\epsilon \rightarrow 0^+}\atop {R\rightarrow +\infty }}\left[\frac{\cos (\pi \alpha )}{2\alpha (1-2\alpha )}\int _{\epsilon }^Re^{-t} t^{1-2\alpha }\,dt +O(R^{-2\alpha })+O(\epsilon ^{2-2\alpha })\right]\\&&\qquad =\frac{\cos (\pi \alpha )\,\Gamma (2-2\alpha )}{2\alpha (1-2\alpha )}=- \cos (\pi \alpha )\,\Gamma (-2\alpha ).\) 
By inserting this information and (REF )
into (REF ), we thereby obtain the desired result in (REF ).

A shorter (but less elementary) proof of (REF )
From Ramanujan's Master Theorem (see e.g.
Formula (B) in Section 11.2 on page 186 of [11]} or
Theorem 3.2 in [12]}),
if a complex-valued function \(f\) 
has an expansion of the form
\( f(x)=\sum _{k=0}^{+\infty }{\frac{\,\ell (k)\,}{k!}}(-x)^{k},\) 
then
\(\int _{0}^{+\infty }x^{s-1}f(x)dx=\Gamma (s)\,\ell (-s).\) 
We take \(s:=1-2\alpha \) 
and \(\ell (s):= \sin \left( \frac{\pi s}{2}\right)\) .
In this way,
\( \ell (k)={\left\lbrace \begin{array}{ll}0 & {\mbox{ if $k\in 2\mathbb {N}$,}}\cr 1& {\mbox{ if $k\in 4\mathbb {N}+1$,}}\cr -1 & {\mbox{ if $k\in 4\mathbb {N}+3$,}}\end{array}\right.}\) 
thus
\( f(x)=-\sum _{j=0}^{+\infty }{\frac{1}{(4j+1)!}}x^{4j+1}+\sum _{j=0}^{+\infty }{\frac{1}{(4j+3)!}}x^{4j+3}=-\sum _{m=0}^{+\infty }{\frac{(-1)^m}{(2m+1)!}}x^{2m+1}=-\sin x.\) 
Then, we deduce from (REF ) that
\(-\int _{0}^{+\infty } x^{-2\alpha }\sin x\,dx=\Gamma (1-2\alpha )\,\sin \left(\frac{\pi }{2}(1-2\alpha )\right)=-2\alpha \Gamma (-2\alpha )\,\cos (\pi \alpha ).\) 
This and (REF ) entail (REF ).
